LeetCode&&劍指offer

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2021-08-03 21:15:22 星期二

509. 斐波那契數

https://leetcode-cn.com/problems/fibonacci-number/submissions/

class Solution {
    public int fib(int n) {
        if(n < 2){
            return n;
        }
        int f = 0, s = 1;
        for(int i = 2; i <= n; i++){
            int t = f + s;
            f = s;
            s = t;
        }
        return s;
    }
}

JZ50 數組中重複的數字

https://www.nowcoder.com/practice/6fe361ede7e54db1b84adc81d09d8524?tpId=13&tqId=11203&tab=answerKey&from=cyc_github
大佬總結:
https://github.com/CyC2018/CS-Notes/blob/master/notes/3. 數組中重複的數字.md

import java.util.*;


public class Solution {
    /**
     * 代碼中的類名、方法名、參數名已經指定，請勿修改，直接返回方法規定的值便可
     *
     * 
     * @param numbers int整型一維數組 
     * @return int整型
     */
	 
	 //方法一
    public int duplicate (int[] numbers) {
        int[] tmp = new int[numbers.length];
        for(int i = 0; i < numbers.length; i++){
            if(tmp[numbers[i]] > 0){
                return numbers[i];
            }
            tmp[numbers[i]]++;
        }
        return -1;
    }
	
	//方法二
 public int duplicate (int[] numbers) {
        for(int i = 0; i < numbers.length; i++){
            while(i != numbers[i]){
                if(numbers[i] == numbers[numbers[i]]){
                    return numbers[i];
                }
                int t = numbers[i];
                numbers[i] = numbers[numbers[i]];
                numbers[t] = t;
            }
        }
         return -1;
     }
}

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2021-08-04 21:32:21 星期三

167. 兩數之和 II - 輸入有序數組

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int i = 0, j = numbers.length - 1;
        while (i < j){
            if (numbers[i] + numbers[j] == target) return new int[] {i + 1, j + 1};
            if (numbers[i] + numbers[j] > target){
                j--;
                continue;
            }else {
                i++;
                continue;
            }
        }
        return null;
    }
}

633. 平方數之和

class Solution {
    public boolean judgeSquareSum(int c) {
        int sqrt = (int) Math.sqrt(c);
        int i = 0, j = sqrt;
        while (i <= j) {
            if (i*i + j*j == c) return true;
            if (i*i + j*j > c){
                j--;
            }else {
                i++;
            }
        }
        return false;
    }
}

345. 反轉字符串中的元音字母

class Solution {
    public String reverseVowels(String s) {
        List<Character> t = Arrays.asList('a', 'o', 'e', 'i', 'u', 'A', 'O', 'E', 'I' ,'U');
        char[] array = s.toCharArray();
        int i = 0, j = array.length - 1;
        while (i <= j){
            if (t.contains(array[i]) && t.contains(array[j])){
                char tmp = array[i];
                array[i] = array[j];
                array[j] = tmp;
                i++;
                j--;
            }else if (t.contains(array[i])){
                j--;
            }else if (t.contains(array[j])){
                i++;
            }else {
                i++;
                j--;
            }
        }
        return new String(array);
    }
}

2021-08-05 23:14:32 星期四

680. 驗證迴文字符串 Ⅱ

class Solution {
    public boolean validPalindrome(String s) {
        if (s == null) return false;
        if (s.length() < 2) return true;
        char[] array = s.toCharArray();
        int i = 0, j = array.length - 1;
        while (i < j) {
            if (array[i] == array[j]) {
                i++;
                j--;
                continue;
            } else {
                return isPalindrome(new String(array, i, j - i)) || isPalindrome(new String(array, i + 1, j - i));
            }
        }
        return true;
    }

    public boolean isPalindrome(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) return false;
            i++;
            j--;
        }
        return true;
    }
}

88. 合併兩個有序數組

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1, j = n - 1, k = m + n - 1;
        while (i >= 0 && j >= 0){
            if (nums1[i] >= nums2[j]){
                nums1[k--] = nums1[i--];
            }else {
                nums1[k--] = nums2[j--];
            }
        }
        if (i >= 0){
            while (k >= 0){
                nums1[k--] = nums1[i--];
            }
        }else {
            while (k >= 0){
                nums1[k--] = nums2[j--];
            }
        }
    }
}

141. 環形鏈表

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) return false;
        ListNode h1 = head, h2 = head.next;
        while (h1 != null && h2 != null && h2.next != null){
            if (h1 == h2) return true;
            h1 = h1.next;
            h2 = h2.next.next;
        }
        return false;
    }
}

2021-08-07 22:49:23 星期六

524. 經過刪除字母匹配到字典裏最長單詞

class Solution {
    public String findLongestWord(String s, List<String> dictionary) {
        if(dictionary == null) return "";
        Collections.sort(dictionary);
        int i = 0, max = 0;
        String result = "";
        while (i < dictionary.size()){
            if (isSubStr(s, dictionary.get(i)) && dictionary.get(i).length() > max){
                max = dictionary.get(i).length();
                result = dictionary.get(i);
            }
            i++;
        }
        return result;
    }
    private boolean isSubStr(String s, String target){
        int i = 0, j = 0;
        while (i < s.length() && j < target.length()){
            if (s.charAt(i) == target.charAt(j)){
                j++;
            }
            i++;
        }
        return j == target.length();
    }
}

215. 數組中的第K個最大元素

class Solution {
    public int findKthLargest(int[] nums, int k) {
        Arrays.sort(nums);
        return nums[nums.length - k];
    }
}

2021-08-08 10:57:32 星期日

20. 有效的括號

class Solution {
    public boolean isValid(String s) {
        char[] array = s.toCharArray();
        Stack<Character> stack = new Stack<>();
        for (char c : array) {
            if (c == '(' || c == '[' || c == '{'){
                stack.push(c);
            }else {
                if (stack.isEmpty()) return false;
                switch (stack.pop()){
                    case '(':
                        if (c != ')') {
                            return false;
                        }
                        break;
                    case '[':
                        if (c != ']'){
                            return false;
                        }
                        break;
                    case '{':
                        if (c != '}'){
                            return false;
                        }
                        break;
                }
            }
        }
        return stack.isEmpty();
    }
}

2021-08-09 21:10:23 星期一

232. 用棧實現隊列

class MyQueue {

    Stack<Integer> inStack = new Stack<>();
    Stack<Integer> outStack = new Stack<>();
    /** Initialize your data structure here. */
    public MyQueue() {

    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        inStack.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if (!outStack.isEmpty()) {
            return outStack.pop();
        }
        while (!inStack.isEmpty()) {
            outStack.push(inStack.pop());
        }
        return outStack.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        if (!outStack.isEmpty()) {
            return outStack.peek();
        }
        while (!inStack.isEmpty()) {
            outStack.push(inStack.pop());
        }
        return outStack.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return outStack.isEmpty() && inStack.isEmpty();
    }
}

69. x 的平方根

class Solution {
    public int mySqrt(int x) {
        if (x < 2) return x;
        int left = 1, right = x;
        while (left <= right){
            int med = (left + right)/ 2;
            int sqrt = x / med;
            if (sqrt == med){
                return med;
            }else if (sqrt > med){
                left = med + 1;
            }else {
                right = med - 1;
            }
        }
        return right;
    }
}