LeetCode-Easy
1.Valid Parenthesesjava
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Input: "()"
Output: true
Input: "([)]"
Output: false
Input: "{[]}"
Output: true
複製代碼
class Solution {
public boolean isValid(String s) {
if(s==null || s.length()%2!=0){
return false;
}
String r = "";
for(int i=0;i<s.length();i++){
char curr = s.charAt(i);
//檢查當前位置字符是 括號的開頭仍是結尾
if(curr == ')' || curr == ']' || curr == '}'){
//若是當前位置字符是 括號的結尾
//檢查拼接的字符串的上一個字符是否可對應消除
if(r == ""){
//若是拼接的字符串爲空:錯
return false;
}else if(curr == ')'&&r.charAt(r.length()-1)=='('){
r = r.substring(0,r.length()-1);
}else if(curr == ']'&&r.charAt(r.length()-1)=='['){
r = r.substring(0,r.length()-1);
}else if(curr == '}'&&r.charAt(r.length()-1)=='{'){
r = r.substring(0,r.length()-1);
}else{
//若是拼接的字符串的上一個字符不可對應消除:錯
return false;
}
}else{
//若是當前位置 是括號的開頭
//沒法消除,則繼續拼接字符串
r += curr;
}
}
//最後檢查拼接的字符串長度,長度!=0,說明有括號未對應消除掉:錯
return r.length()==0;
}
}
class Solution {
public boolean isValid(String s) {
if(s.length() != 0 && s.length()%2 != 0){
return false;
}
char[] c = new char[s.length()];
char curr;
int length = 0;
for(int i = 0;i<s.length();i++){
curr = s.charAt(i);
if(length > 0 && (curr==')'||curr==']'||curr=='}') && match(curr,c[length-1])){
length --;
}else{
c[length] = curr;
length++;
}
}
return length == 0;
}
public boolean match(char curr,char pre){
if(curr == ')' && pre == '('){
return true;
}
if(curr == ']' && pre == '['){
return true;
}
if(curr == '}' && pre == '{'){
return true;
}
return false;
}
}
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2.N-Repeated Element in Size 2N Arraynode
In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.
Return the element repeated N times.
Input: [1,2,3,3]
Output: 3
Input: [2,1,2,5,3,2]
Output: 2
複製代碼
思路:
1:
既然有1個數字佔據了數組的1/2,咱們將數組進行分割,2個元素1組,極端狀況下用於看不到重複元素.
可是若是3個元素一組, 即便是極端狀況,咱們也能夠在某一組3個元素中檢查到重複對象,
若是檢查不到,分組剩下的元素就必定是目標值.
2:
固然能夠利用現成的API,使用Set.add每一個元素,經過返回的boolean值來判斷是否取到重複元素.
class Solution {
public int repeatedNTimes(int[] A) {
int count = A.length/3;
for(int i=0;i<count;i++){
if(A[i*3] == A[i*3 + 1] || A[i*3] == A[i*3 + 2]) return A[i*3];
if(A[i*3 + 1] == A[i*3 + 2]) return A[i*3 + 1];
}
return A[3*count];
}
}
複製代碼
3.Remove Elementgit
Given an array nums and a value val, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length. Example 1: Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length. 複製代碼
class Solution {
public int removeElement(int[] nums, int val) {
if(nums==null||nums.length==0) return 0;
//首先定義返回值length,默認爲0
int length = 0;
for(int i=0;i<nums.length;i++){
//若是遍歷元素和給定值不一樣,則將length對應的索引指向當前元素,而後length+1
if(nums[i]!=val) nums[length++]=nums[i];
}
return length;
}
}
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4.Merge Two Sorted Lists數組
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
複製代碼
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null || l2 == null) return l1==null ? l2 : l1;
ListNode first = l1.val < l2.val ? l1 : l2;
ListNode second = first == l1 ? l2 : l1;
first.next = mergeTwoLists(first.next,second);
return first;
}
}
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5.Remove Duplicates from Sorted Arraybash
Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond the returned length. Example 2: Given nums = [0,0,1,1,1,2,2,3,3,4], Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.
複製代碼
思路:
既然是已經排好序的數組去除重複值,能夠讓 下一個元素/j 和 當前元素/i 做比較:
若是二者相等,則返回值不變.
若是二者不相等,將當前返回值索引指向當前元素/j,並將返回值+1
每次遍歷,i和j都+1
class Solution {
public int removeDuplicates(int[] nums) {
if(nums==null||nums.length==0) return 0;
if(nums.length==1) return 1;
int length = 1;
int i = 0;
for(int j = 1;j<nums.length;j++){
if(nums[j]!=nums[i]){
nums[length] = nums[j];
length++;
}
i++;
}
return length;
}
}
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6.Two Sumapp
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
複製代碼
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
for(int i=0;i<nums.length;i++){
Integer targetKey = target - nums[i];
if(map.containsKey(targetKey)){
return new int[]{map.get(targetKey),i};
}
map.put(nums[i],i);
}
return null;
}
}
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7.Reverse Integeride
Given a 32-bit signed integer, reverse digits of an integer.
Input: 123
Output: 321
Input: -123
Output: -321
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
複製代碼
class Solution {
public int reverse(int x) {
int result = 0;
while(x != 0){
int tail = x % 10;
x /= 10;
if(result > Integer.MAX_VALUE / 10 || (result == Integer.MAX_VALUE / 10 && tail > 7)){
return 0;
}
if(result < Integer.MIN_VALUE / 10 || (result == Integer.MIN_VALUE / 10 && tail < -8)){
return 0;
}
result = result * 10 + tail;
}
return result;
}
}
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8.Palindrome Numberui
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Input: 121
Output: true
Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
複製代碼
思路1:將整數從中間分隔,而後依次比較'首尾對應的2個數字'是否相等,出現不等說明不是迴文數
class Solution {
public boolean isPalindrome(int x) {
if(x < 0){
return false;
}
if(x < 10){
return true;
}
int length = (int)(Math.log10(x) + 1);
for(int i = 0; i < length/2; i++){
if(x/((int)Math.pow(10,i))%10 != x/((int)Math.pow(10,length-1-i))%10){
return false;
}
}
return true;
}
}
思路2:將整數直接反轉,而後比較反轉後數字和元數字是否相等,相等即爲迴文數
class Solution {
public boolean isPalindrome(int x) {
if(x<0 || (x!=0 && x%10==0)) return false;
// if(x<10) return true;
int reverse = 0;
while(x > reverse){
reverse = reverse * 10 + x % 10;
x /= 10;
}
return x==reverse || x==reverse/10;
}
}
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9.Roman to Integerthis
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II. Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: I can be placed before V (5) and X (10) to make 4 and 9. X can be placed before L (50) and C (100) to make 40 and 90. C can be placed before D (500) and M (1000) to make 400 and 900. Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. Input: "III" Output: 3 Input: "IV" Output: 4 Input: "IX" Output: 9 Input: "LVIII" Output: 58 Explanation: L = 50, V= 5, III = 3. Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. 複製代碼
class Solution {
public int romanToInt(String s) {
/* I 1 V 5 X 10 L 50 C 100 D 500 M 1000 */
int[] nums = new int[s.length()];
for(int i=0;i<nums.length;i++){
if(s.charAt(i)=='I'){
nums[i]=1;
}else if(s.charAt(i)=='V'){
nums[i]=5;
}else if(s.charAt(i)=='X'){
nums[i]=10;
}else if(s.charAt(i)=='L'){
nums[i]=50;
}else if(s.charAt(i)=='C'){
nums[i]=100;
}else if(s.charAt(i)=='D'){
nums[i]=500;
}else if(s.charAt(i)=='M'){
nums[i]=1000;
}
}
int result = 0;
for(int i=0;i<nums.length-1;i++){
if(nums[i]<nums[i+1]){
result -= nums[i];
}else{
result += nums[i];
}
}
result += nums[nums.length - 1];
return result;
}
}
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10.Longest Common Prefixspa
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".
Input: ["flower","flow","flight"]
Output: "fl"
Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Note:
All given inputs are in lowercase letters a-z.
複製代碼
class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs==null||strs.length==0){
return "";
}
StringBuilder sb = new StringBuilder("");
for(int i=0;i<strs[0].length();i++){
for(int j=1;j<strs.length;j++){
if(strs[j].length()<(i+1) || strs[j].charAt(i) != strs[0].charAt(i)){
return sb.toString();
}
}
sb.append(strs[0].charAt(i));
}
return sb.toString();
}
}
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11.Implement strStr()
Implement strStr().
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Input: haystack = "hello", needle = "ll"
Output: 2
Input: haystack = "aaaaa", needle = "bba"
Output: -1
Clarification:
What should we return when needle is an empty string? This is a great question to ask during an interview.
For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
複製代碼
class Solution {
public int strStr(String haystack, String needle) {
if(needle==null||needle.equals("")) return 0;
if(needle.length()>haystack.length() || (needle.length()==haystack.length()&&!needle.equals(haystack))) return -1;
for(int j=0;j<=haystack.length()-needle.length();j++){
for(int i=0;i<needle.length();i++){
if(needle.charAt(i) != haystack.charAt(j+i)) break;
if(i == needle.length() - 1){
return j;
}
}
}
return -1;
}
}
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12.Search Insert Position
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Input: [1,3,5,6], 5
Output: 2
Input: [1,3,5,6], 2
Output: 1
Input: [1,3,5,6], 7
Output: 4
Input: [1,3,5,6], 0
Output: 0
複製代碼
class Solution {
public int searchInsert(int[] nums, int target) {
if(target > nums[nums.length - 1]){
return nums.length;
}
if(target < nums[0]){
return 0;
}
for(int i = 0; i< nums.length; i++){
if(target <= nums[i]){
return i;
}
}
return 0;
}
}
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13.Search Insert Position
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