Hackerrank Practice

時間 2020-01-13

標籤 hackerrank practice 简体版

原文原文鏈接

Anagram 拆分數組看一半操做幾回能成爲另外一半的anagram

題目

輸入第一行是要判斷的字符串的個數n，以後的n行輸入爲須要判斷的字符串。
每一個字符串str，是兩個等長字符串的合體，因此長度必定是偶數。若爲奇數，返回-1。
因此str分紅兩個substring，右邊須要多少次操做能變爲左邊。只要用一個26位的數組統計每一個字符在a和b中出現的次數就行，在a中出現+1，在b中出現-1，最後統計數組的26位元素的絕對值除以2，就是要進行操做的次數。java

 Input Formatnode

The first line will contain an integer T representing the number of test cases. Each test case will contain a string having length (a+b) which will be concatenation of both the strings described in problem. The string will only contain small letters and without any spaces.git

 Output Formatexpress

An integer corresponding to each test case is printed in a different line i.e., the number of changes required for each test case. Print ‘-1’ if it is not possible.
 Constraints
1 ≤ T ≤ 100  1 ≤ a+b ≤ 10,000
 
Sample Input數組

5
aaabbb
ab
abc
mnop
xyyx

Sample Outputapp

 Explanationless

In the five test caseside

One string must be 「aaa」 and the other 「bbb」. The lengths are a=3 and b=3, so the difference is less than 1. No characters are common between the strings, so all three must be changed.ui
One string must be 「a」 and the second 「b」. The lengths are a=1 and b=1, so the difference is less than 1. One character must be changed to them the same.this
Since the string lengths a and b must differ by no more than 1, the lengths are either a=1 and b=2 or a=2 and b=1.No sequence of substitutions will make the two anagrams of one another.
One string must be 「mn" and other be 「op」. The length are a=2 and b=2, so the difference is less than 1. Nocharacters are common between the strings, so both must be changed.
One string must be 「xy」 and the other be 「yx」. The length are a=2 and b=2, so the difference is less than 1. No changes are needed because the second string is already an anagram of the first.
Collapse Question

import java.io.*;
import java.util.*;
public class Solution {
    public static void main(String args[] ) throws Exception {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
        Scanner in = new Scanner(System.in);
        String str = in.nextLine();
        int num = Integer.parseInt(str);
        String[] strs = new String[num];
        for (int i = 0; i < num; i++) {
            strs[i] = in.nextLine();
        }
        for (String s: strs) {
            int res = validate(s);
            System.out.println(res);
        }
    }

public static int validate(String str) {
    if (str.length() % 2 == 1) return -1;
    int len = str.length();
    String a = str.substring(0, len/2);
    String b = str.substring(len/2);
    int[] ch = new int[26];
    for (int i = 0; i < a.length(); i++) {
        char ca = a.charAt(i);
        char cb = b.charAt(i);
        if (ca < 'a' || ca > 'z' || cb < 'a' || cb > 'z') return -1;
        ch[ca-'a']++;
        ch[cb-'a']--;
    }
    int count = 0;
    for (int c: ch) {
        if (c == 0) continue;
        count += Math.abs(c);
    }
    return count/2;
}
}

Count Duplicates 數組中重複元素的個數

題目

找到一個數組中重複元素的個數。

Complete the countDuplicates function in the editor below. It has 1 parameter: an array of integers, numbers. It must return an integer denoting the number of non-unique values in the numbers array.

Constraints

1 ≤ n ≤ 1000
1 ≤ numbersi ≤ 1000

Solution

static int countDuplicates(int[] numbers) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int key: numbers) {
            if (map.get(key) == null) map.put(key, 1);
            else map.put(key, map.get(key)+1);
            //map.put(key, map.getOrDefault(key, 0)+1);
        }
        int count = 0;
        for (Map.Entry<Integer, Integer> entry: map.entrySet()) {
            if (entry.getValue() > 1) count++;
        }
        return count;
    }

或者用數組

static int countDuplicates(int[] numbers) {
        int[] n = new int[1000];
        Arrays.fill(n, 0);
        int count = 0;
        for (int i: numbers) {
            if (n[i] == -1) continue;
            else if (n[i] == 0) n[i]++;
            else { // actually n[i] = 1
                n[i] = -1;
                count++;
            }
        }
        return count;
    }

HackLand Election 誰選票多且名字靠後

題目

n我的投票，名字（票數）最多的獲勝，若票數相同，名字字母順序靠後的獲勝。

Solution

static String electionWinner(String[] votes) {
        Map<String, Integer> map = new HashMap<>();
        int max = 0;
        for (String vote: votes) {
            if (map.get(vote) == null) map.put(vote, 1);
            else {
                int count = map.get(vote)+1;
                map.put(vote, count);
                max = Math.max(max, count);
            }
            //map.put(vote, map.getOrDefault(vote, 0)+1);
        }
        List<String> res = new ArrayList<>();
        for (Map.Entry<String, Integer> entry: map.entrySet()) {
            if (entry.getValue() == max) res.add(entry.getKey());
        }
        Collections.sort(res);
        return res.get(res.size()-1);
    }

Delete Nodes Greater Than X 鏈表按值刪除結點

題目

從鏈表裏刪除全部值大於X的結點。
須要dummy結點，用cur.next.val去比大小。

Solution

static LinkedListNode removeNodes(LinkedListNode list, int x) {
        LinkedListNode dummy = new LinkedListNode(0);
        dummy.next = list;
        LinkedListNode cur = dummy;
        while (cur.next != null) {
            if (cur.next.val > x) {
                cur.next = cur.next.next;
            }
            else cur = cur.next;
        }
        return dummy.next;
    }

Even Odd Query 數組裏取x,y 看pow(x,y)奇偶性

題目

其實就是給一個數組，而後給一些index pair，看數組中對應index的兩個數的pow(x, y)是奇數仍是偶數。

Input Format

 The first line contains an integer N.
The next line contains N space separated single-digit integers (whole numbers between 0 and 9). The third line contains a positive integer Q, denoting the number of queries to follow.
Each of the subsequent Q lines contains two positive integers x and y separated by a single space.

Output Format

 For each query, print the "Even" if the value returned is even, otherwise print "Odd " without quotes.

Sample Input #00

Sample Output #00

Odd
Even

Explanation #00

find(1,2) = 3^2 = 9, which is odd.
 find(2,3) = 2^7 = 128, which is even.

Solution

import java.io.*;
import java.util.*;
public class Solution {
    public static void main(String args[] ) throws Exception {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
        //BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        //int N = Integer.parseInt(br.readLine());
        StringBuffer sb = new StringBuffer();
        Scanner in = new Scanner(System.in);
        int N = Integer.parseInt(in.nextLine());
        int[] A = new int[N];
        String[] strs = in.nextLine().split(" ");
        for (int i = 0; i < N; i++) {
            A[i] = Integer.parseInt(strs[i]);
        }
        for (int i = Integer.parseInt(in.nextLine()); i > 0; i--) {
            strs = in.nextLine().split(" ");
            int m = Integer.parseInt(strs[0])-1;
            int n = Integer.parseInt(strs[1])-1;
            int base = A[m];
            int power = (m == n) ? 1 : A[m+1];
            if (power == 0 || (base & 1) == 1) sb.append("Odd\n");
            else sb.append("Even\n");
        }
        System.out.print(sb);
    }
}

The Bit Game 在某區間內最大的兩數XOR值

題目

在l和r之間取兩個數x，y，使他倆的異或值<=k且最大，返回這個值。

Solution

static int maxXor(int l, int r, int k) {
        if (l == r) return 0;
        if (l > r) return maxXor(r, l, k);
        int max = 0;
        for (int i = l; i <= r; i++) {
            for (int j = l; j <= r; j++) {
                if ((i ^ j) > k) continue;
                max = Math.max(max, i ^ j);
            }
        }
        return max;
    }

Element Present in BST Tree 在BST裏找元素

題目

看一個BST裏有沒有值爲val的元素，有返回1，沒有返回0。
用cur去遍歷，用pre存儲cur在本次循環起始的位置，誰知道cur.left和cur.right存在不存在呢，對吧。

Solution

private static int isPresent(Node root, int val){
        Node pre = root, cur = root;
        while (cur != null) {
            pre = cur;
            if (cur.val == val) return 1;
            else if (cur.val < val) cur = cur.right;
            else cur = cur.left;
        }
        if ((pre.left != null && pre.left.val == val) || (pre.right != null && pre.right.val == val)) return 1;
        return 0;
    }

Pangrams

題目

給一些字符串，看是否是pangram，即包含所有26個字母的句子。是就返回1，不是就返回0.

Solution

static String isPangram(String[] strings) {
        StringBuilder sb = new StringBuilder();
        for (String str: strings) {
            int[] pan = new int[26];
            Arrays.fill(pan, 0);
            for (int i = 0; i < str.length(); i++) {
                char ch = str.charAt(i);
                if (ch >= 'a' && ch <= 'z') pan[ch-'a']++;
            }
            for (int i = 0; i < 26; i++) {
                if (pan[i] == 0) {
                    sb.append("0");
                    break;
                }
                if (i == 25 && pan[i] != 0) sb.append("1");
            }
        }
        return sb.toString();
    }

Cut the Sticks 每次按最短那根的長度切全部木頭，每次切完記錄剩餘的木頭數

題目

有N條木段，每次切割必須這樣：把全部的木段都切掉最短的那條木段的長度。每次切割後記錄剩餘木段的條數，最短的木段確定沒了嘛，如此重複，直到都被切禿嚕了（爲0）。

Sample Case:

lengths       cut length   count cuts
1 2 3 4 3 3 2 1       1           8
_ 1 2 3 2 2 1 _       1           6
_ _ 1 2 1 1 _ _       1           4
_ _ _ 1 _ _ _ _       1           1
_ _ _ _ _ _ _ _     DONE        DONE

Solution

static int[] cutSticks(int[] lengths) {
        int n = lengths.length;
        Arrays.sort(lengths);
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (lengths[i] == 0) continue;
            else {
                ans.add(n-i);
                int temp = lengths[i];
                for (int j = i; j < n; j++) {
                    lengths[j] -= temp;
                }
            }
        }
        int[] res = new int[ans.size()];
        for (int i = 0; i < ans.size(); i++) res[i] = ans.get(i);
        return res;
    }

Prime or Not? 判斷質數返回最大因數

題目

是質數，返回1；不是質數，返回最大因數。

Solution

static int isPrime(long n) {
        if (n <= 2) return 1;
        else if (n > 2 && n % 2 == 0) return 2;
        for (int i = 2; i <= n/2; i++) {
            if (n % i == 0) return i;
        }
        return 1;
    }

Reduce the Fraction 約分約分

解法

找最大公約數，分子和分母都除以公約數後加入StringBuilder，再添加到結果數組便可。

static String[] ReduceFraction(String[] fractions) {
        String[] res = new String[fractions.length];
        for (int i = 0; i < fractions.length; i++) {
            String[] nums = fractions[i].split("/");
            int[] n = new int[2];
            n[0] = Integer.parseInt(nums[0]);
            n[1] = Integer.parseInt(nums[1]);
            
            StringBuilder sb = new StringBuilder();
            sb.append(n[0]/(GCD(n[0], n[1]))+"/"+n[1]/(GCD(n[0], n[1])));
            res[i] = sb.toString();
        }
        return res;
    }
    static int GCD(int n1, int n2) {
        if (n2 == 0) return n1;
        return GCD(n2, n1%n2);
    }

Pascal's Triangle 帕斯卡三角

Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
....

題目

給出層數n，print出帕斯卡三角形。

Solution

static void pascalTriangle(int k) {
        for (int i = 0; i < k; i++) {
            for (int j = 0; j <= i; j++) {
                System.out.print(pascal(i, j) + " ");
            }
            System.out.println();
        }

    }
    static int pascal(int i, int j) {
        if (j == 0) return 1;
        if (i == j) return 1;
        return pascal(i-1, j-1) + pascal(i-1, j);
    }

Calculate Factorial 破階乘

n前面要用(long)哦。

Solution

static long factorial(int n) {
        long res = 1;
        if (n <= 0) return 0;
        while (n != 0) {
            res *= (long)n;
            n--;
        }
        return res;
    }

Angry Children 給小孩發糖果令分配公平不懸殊

題目

N包糖，裏面的糖果數不一樣，挑k包，分給k個孩子，要求找到最小的分配糖果差值。
Sample Input #2

4 //children
10 //packet #
1 2 3 4 10 20 30 40 100 200

Sample Output #2

3

Explanation #2

 Here K = 4. We can choose the packets that contain 1,2,3,4 candies.
The unfairness is max(1,2,3,4) - min(1,2,3,4) = 4 - 1 = 3

Solution

static int minUnfairness(int k, int[] arr) {
        //actually find the min of arr[i+k-1]-arr[i]
        Arrays.sort(arr);
        int n = arr.length;
        int min = arr[n-1];
        for (int i = 0; i < n-k+1; i++) {
            min = Math.min(min, arr[i+k-1]-arr[i]);
        }
        return min;
    }

Is Possible AKA Can Reach (a,b)可否成爲(c,d)

題目

Consider a pair of integers, (a, b). We can perform the following operations on (a, b) in any order, zero or more times:

(a, b) → (a + b, b)
(a, b) → (a, a + b)

Solution

c, d互減，直到c <= a && d <= b的時候。
若此時c == a && d == b，說明a, b的確能夠變成c, d.

static String isPossible(int a, int b, int c, int d) {
        while (c > a || d > b) {
            if (c > d) {
                c -= d;
                if (c < a) return "No";
            }
            else {
                d -= c;
                if (d < b) return "No";
            }
        }
        if (a == c && b == d) return "Yes";
        else return "No";
    }

Encircular 旋轉的機器人可否回到咱們的原點

G instructs the robot to move forward one step.
L instructs the robot to turn left.
R instructs the robot to turn right.

題目

循環執行轉彎、前進的一串動做，可否回到原點？告訴你，只和角度有關。一個L，就是90°，再來一個R，就把它抵消了，那就回不去了。最後剩下的L爲奇數個，就回得去。

Solution

static String[] doesCircleExist(String[] commands) {
        int len = commands.length;
        String[] res = new String[len];
        for (int i = 0; i < len; i++) {
            res[i] = valid(commands[i]);
        }
        return res;
    }
    static String valid(String com) {
        int l = 0;
        for (int i = 0; i < com.length(); i++) {
            char ch = com.charAt(i);
            if (ch == 'G') continue;
            if (ch == 'L') l++;
            if (ch == 'R') l--;
        }
        if (l % 2 == 0) return "NO";
        else return "YES";
    }

Circle

題目

給一堆齒輪找各自符合組合標準且造價最低的那個齒輪的index。
循環，每一個齒輪，找合適的另外一個齒輪分兩步：先找到符合尺寸標準的全部齒輪，再從這些齒輪中找到造價最低的那個，將index加入結果數組。

Solution

static int[] Circles(int distance, int[] radius, int[] cost) {
        int[] result = new int[radius.length];
        for (int i = 0; i < radius.length; i++) {
            List<Integer> tmp = new ArrayList<Integer>();
            for (int j = 0; j < radius.length; j++) {
                if (radius[j] >= distance - radius[i]) tmp.add(j);//添加知足尺碼的index
            }
            result[i] = getSmallest(cost, tmp, i);
        }
        return result;
    }

    static int getSmallest(int[] cost, List<Integer> tmp, int i) {
        if (tmp.size() == 0) return 0;
        int index = tmp.get(0);
        int mincost = cost[tmp.get(0)];
        for (int j = 1; j < tmp.size(); j++) {
            if (cost[tmp.get(j)] < mincost) {
                mincost = cost[tmp.get(j)];
                index = tmp.get(j);
            }
        }
        return index+1;
    }

List Max 著名的Prefix Sum

題目

不斷給某個區間作加同一個數的操做，問最後最大的數變成了多少。

Sample Input 0
5 3
1 2 100
2 5 100
3 4 100

Sample Output 0
200

Explanation 0
We perform the following sequence of m = 3 operations on list = {0, 0, 0, 0, 0}:

Add k = 100 to every element in the inclusive range [1, 2], resulting in list = {100, 100, 0, 0, 0}.
Add k = 100 to every element in the inclusive range [2, 5], resulting in list = {100, 200, 100, 100, 100}.
Add k = 100 to every element in the inclusive range [3, 4], resulting in list = {100, 200, 200, 200, 100}.
We then print the maximum value in the final list, 200, as our answer.

import java.io.*;
import java.util.*;
public class Solution {
    public static void main(String args[] ) throws Exception {
        Scanner in = new Scanner(System.in);
        int size = in.nextInt(); //size of array
        int m = in.nextInt(); //# of operations
        //marking array
        long[] arr = new long[size+1]; //allocate one more space for the last digit to sum up
        for (int i = 0; i < m; i++) {
            int a = in.nextInt();
            int b = in.nextInt();
            int k = in.nextInt();
            arr[a-1] += k; //only mark the starting index that changed
            arr[b] -= k; //從這日後的最後都會被加上以前的sum，因此提早減掉 
        }
        long max = Long.MIN_VALUE; //用long哦  
        long sum = 0;
        for (int i = 0; i < size; i++) {
            sum += arr[i];
            max = Math.max(max, sum);
        }
        System.out.println(max);
    }
}

Number Complement 找一個數的補碼

找到最高位，創建掩碼mark取反

static int getIntegerComplement(int n) {
        int high = (int)(Math.log(n)/Math.log(2));
        int mark = (1 << (high+1))-1;
        return n ^ mark;
    }

The Love-Letter Mystery

題目

letter的每個句子是一個字符串，咱們要計算將這些字符串轉變爲迴文字符串的操做次數。
例如a-->c，要走兩次哦。那麼"abc"的步數就是2，"abcd"的步數就是3+1=4.

static int[] mystery(String[] letter) {
        if (letter == null || letter.length == 0) return new int[0];
        if (letter.length == 1) {
            int[] a = new int[1];
            a[0] = 0;
            return a;
        }
        int[] res = new int[letter.length];
        for (int i = 0; i < letter.length; i++) {
            String str = letter[i];
            int op = helper(str);
            res[i] = op;
        }
        return res;
    }
    static int helper(String str) {
        int n = str.length();
        int left = 0, right = n-1;
        int count = 0;
        while (left < right) {
            count += Math.abs(str.charAt(right) - str.charAt(left));
            left++;
            right--;
        }
        return count;
    }

Metals 章魚賣鐵咋切最賺

static int maxProfit(int cost_per_cut, int metal_price, int[] lengths) {
        int maxLength = 0;
        //找最長的鐵塊
        for (int length : lengths) {
            if (length > maxLength) {
                maxLength = length;
            }
        }
        int maxProfit = 0;
        for (int i = 1; i < maxLength; i++) {
            int sumOfLengths = 0;
            int sumOfCutCounts = 0;
            int sumOfCutWastes = 0;
            for (int length : lengths) {
                sumOfLengths += length;
                if (length % i == 0) {
                    sumOfCutCounts += (length/i - 1); //總長被定長整除，能夠少切一次
                } else {
                    sumOfCutCounts += (length/i);
                }
                sumOfCutWastes += (length%i); //統計浪費的零頭
            }
            int profit = sumOfLengths*metal_price - sumOfCutCounts*cost_per_cut - sumOfCutWastes*metal_price; //總價 - 切割費用 - 浪費材料價值
            if (profit > maxProfit) {
                maxProfit  = profit; //更新最大利潤
            }
        }
        return maxProfit;
    }


#Gem Stones 有幾個字符在全部字符串都出現了呢

static int gemstones(String[] rocks) {
    Set<Character> set = new HashSet<>(26);            
    Set<Character> cur = new HashSet<>(26);
    char[] init = rocks[0].toCharArray();
    for (char ch: init) set.add(ch);
    for (String rock: rocks) {
        for (char ch: rock.toCharArray()) {
            cur.add(Character.valueOf(ch));
        }
        set.retainAll(cur);
        cur.clear();
    }
    return set.size();
}

Valid BST using Pre-traversal

import java.io.*;
import java.util.*;

public class Solution {

public static void main(String args[] ) throws Exception {
    Scanner in = new Scanner(System.in);
    int n = Integer.parseInt(in.nextLine());
    for (int i = 0; i < n; i++) {
        int len = Integer.parseInt(in.nextLine());
        int[] nodes = new int[len];
        String[] str = in.nextLine().split(" ");
        for (int j = 0; j < len; j++) {
            nodes[j] = Integer.parseInt(str[j]);
        }
        if (validBST(nodes, 0, len-1)) System.out.println("YES");
        else System.out.println("NO");
    }
}
public static boolean validBST(int[] A, int start, int end) {
    if (end <= start) return true;
    int root = A[start];
    int i = start+1;
    while (i <= end && A[i] < root) i++; //第一個大於root的數，只能是右子樹root.right
    int right = i; 
    while (i <= end && A[i] > root) i++;
    if (i != end+1) return false; //右子樹有小於等於root的結點，上面的while提早結束了
    return validBST(A, start+1, right-1) && validBST(A, right, end); //左子樹OK && 右子樹OK
}

}

#Is this a tree? 難題

public static String SExpression(String s){
    boolean[][] graph = new boolean [26][26];
    HashSet<Character> nodes = new HashSet<>();
    //construct graph and check error E2: duplicate edges
    boolean E2 = false;
    for(int i=1;i<s.length();i+=6){
        int x = s.charAt(i)-'A', y = s.charAt(i+2)-'A';
        if(graph[x][y]) //duplicate edge
            E2 = true;
        graph[x][y] = true;
        nodes.add(s.charAt(i));
        nodes.add(s.charAt(i+2));
    }
    //check error E1: more than 2 children
    boolean E1 = false;
    for(int i=0;i<26;i++){
        int count = 0; //number of child
        for(int j=0;j<26;j++){
            if(graph[i][j])
                count++;
        }
        if(count>2)
            return "E1";
    }
    if(E2) return "E2"; //return E2 after checking E1
    
    //check E3: cycle present and E4: multiple roots
    int numOfRoots = 0;
    char root =' ';
    for (char node: nodes){ //only check char that in the tree
        for(int i=0;i<26;i++){
            if(graph[i][node-'A'])
                break;
            if(i==25){
                numOfRoots++;
                root = node;
                boolean[] visited = new boolean[26];
                if(IsCycle(node, graph, visited))
                    return "E3";
            }
        }
    }
    if(numOfRoots==0) return "E3"; //if no root, must be a cycle
    if(numOfRoots>1) return "E4"; //if more than one roots
    if(root==' ') return "E5"; //if no edge in input string, invalid input error
    return GetExpressionHelper(root, graph);
    
}

//true means there is a cycle, false means no cycle
private static boolean IsCycle(char node, boolean[][] graph, boolean[] visited){
    if(visited[node-'A']) //node has already been visited, must has a cycle
        return true;
    visited[node-'A'] = true;
    for(int i=0;i<26;i++){
        if(graph[node-'A'][i]){
            if(IsCycle((char)(i+'A'), graph, visited))
                return true;
        }
    }
    return false;
}

//Recursive DFS to get the expression/construct the tree
private static String GetExpressionHelper(char root, boolean[][] graph){
    String left = "", right = ""; //if no children, left and right should be empty
    for(int i=0;i<26;i++){
        if(graph[root-'A'][i]){
            left = GetExpressionHelper((char)(i+'A'), graph);
            for(int j=i+1;j<26;j++){
                if(graph[root-'A'][j]){
                    right = GetExpressionHelper((char)(j+'A') ,graph);
                    break;
                }
            }
            break;
        }
    }
    return "("+root+left+right+")";
}

#K-Difference 這是Two Sum的變種

static int kDifference(int[] a, int k) {

    int n = a.length;
    int count = 0;
    Arrays.sort(a);
    for(int i=0; i<n-1; i++){
        int j = i+1;
        while(a[j]-a[i] < k) {
            j++;
            if(j==n) break;
        }
        if (j == n) continue;
        if(a[j]-a[i]==k) count++;
    }
    return count;
}

#Maximum Difference in an Array -- Nice

數組中任意兩元素的最大差值，必須是後面的減前面的。

brute force -- TLE

static int maxDifference(int[] a) {
    if (a == null || a.length < 2) return 0;
    int max = -1;
    int n = a.length;
    for (int i = 0; i < n-1; i++) {
        for (int j = i+1; j < n; j++) {
            if (a[j] > a[i]) max = Math.max(max, a[j]-a[i]);
        }
    }
    return max;
}

**此方法精彩**

static int maxDifference(int[] a) {
    if (a == null || a.length < 2) return 0;
    int n = a.length;
    int res = -1;
    for (int i = 0; i < n; i++) {
       int maxIndex = getmax(a, i);
       int minIndex = getmin(a, i, maxIndex);
       if (a[maxIndex]-a[minIndex] > res) res = a[maxIndex]-a[minIndex] > 0 ? a[maxIndex]-a[minIndex] : -1;
       i = maxIndex+1;
    }
    return res;
}
static int getmax(int[] a, int start) {
    int maxIndex = start;
    for (int i = start; i < a.length; i++) {
        if (a[i] > a[maxIndex]) maxIndex = i;
    }
    return maxIndex;
}
static int getmin(int[] a, int start, int end) {
    int minIndex = start;
    for (int i = start+1; i <= end; i++) {
        if (a[i] < a[minIndex]) minIndex = i;
    }
    return minIndex;
}

**此方法更精彩**

static int maxDifference(int[] a) {
    if (a == null || a.length < 2) return 0;
    int n = a.length;
    int res = -1;
    int min = a[0];
    for (int i = 0; i < n; i++) {
        if (a[i] < min) min = a[i];
        else {
            if (a[i]-min > 0) res = Math.max(res, a[i]-min);
            else continue;
        }
    }
    return res;
}

#Sum of Two Numbers in an Array (sum = k)

Your function must return the number of pairs of `unique/distinct` pairs in a having a sum equal to k.

static int numberOfPairs(int[] a, long k) {
    Arrays.sort(a);
    int n = a.length;
    int i = 0, j = n-1;
    int count = 0;
    while (i < j) {
        while (i > 0 && a[i] == a[i-1]) i++; //skip the duplicate
        while (j < n-1 && a[j] == a[j+1]) j--; //skip the duplicate
        long sum = a[i]+a[j];
        if (sum == k) {
            count++;
            i++;
            j--;
        }
        else if (sum < k) i++;
        else j--;
    }
    return count;
}

#Balance the Array 找兩邊sum相等的index

static int balanceSum(int[] A) {
    int[] sum = new int[A.length];
    sum[0] = A[0];
    for (int i = 1; i < A.length; i++) {
        sum[i] = sum[i-1] + A[i];
    }
    for (int i = 1; i < A.length-1; i++) {
        if (sum[i-1] == sum[A.length-1] - sum[i]) return i;
    }
    return -1;
}

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