java代码如下: publicclassSolution {publicintmissingNumber(int[] nums) {intsum = nums.len * (nums.len + 1)/2; //length大小是n-1for(inti = 0; i < nums.length; ++i){ sum-=nums[i]; }returnsum; } }
运用异或就行了。 代码如下: publicclassSolution {publicintmissingNumber(int[] nums) {intlength =nums.length;intres = 0;for(inti = 1; i <= length; i++) res^=i;for(inti = 0; i < length; i++) res^=nums[i];returnres; } }...
因为数字是从0, 1, 2, …, n中抽走一个,因此,该数字的值是0, 1, 2, …, n的和减去现有数字的和。 public class Solution { public int missingNumber(int[] nums) { int sum = 0; int len = nums.length; for(int i = 0; i < len; i++){ sum += nums[i]; } return len * (len...
代码 public class Solution { public int missingNumber(int[] nums) { int res = 0; for(int i = 0; i <= nums.length; i++){ res ^= i == nums.length ? i : i ^ nums[i]; } return res; } } First Missing Positive Given an unsorted integer array, find the first missing positiv...
public class Solution { public int missingNumber(int[] nums) { //首先对数组进行排序 Arrays.sort(nums); int startData=nums[0]; for(int i=1;i<nums.length;i++) { //检查数组是否连续 if((startData+1)==nums[i]) { startData=nums[i]; ...
Is there an existing issue for this? I have searched the existing issues Feature Description Leetcode solution 176 is missing Use Case Benefits No response Add ScreenShots No response Priority High Record I have read the Contributing Gui...
LeetCode Username endlesscheng Problem Number, Title, and Link Maximize the Minimum Game Score https://leetcode.com/problems/maximize-the-minimum-game-score/description/ Bug Category Missing test case (Incorrect/Inefficient Code getting ...
【LeetCode】268. Missing Number 丢失的数字(Easy)(JAVA) 题目地址: https://leetcode.com/problems/missing-number/ 题目描述: Given an array nums containing n distinct numbers in the range [0, n], return the only number ...LeetCode 268. 丢失的数字 Missing Number 给定一个包含 [0, n] 中...
Java class Solution { public int firstMissingPositive(int[] nums) { for (int i = 0; i < nums.length; i++) { /* * 整个算法中最关键的操作就是判断何时需要将当前值交换到对应下标处 * 需要同时满足以下四点: * 1. 当前值为正整数,不然无需处理 ...
classSolution{publicintmissingNumber(int[] nums){intsum=0;intlen=nums.length;for(inti=0; i < len; i++) { sum += nums[i]; }returnlen*(len+1)/2- sum; } } 解法二 思路 用位运算。运用java中的异或来解题。异或性质如下: 1、交换律 ...