classSolution {publicintsingleNonDuplicate(int[] nums) {intl=0,r=nums.length-1;//注意这里r的取值应该为-1while(l<r){intm=l+(r-l)/2;if(m%2==1) m--;if(nums[m]==nums[m+1]) l=m+2;elser=m; }returnnums[l]; } }
Can you solve this real interview question? Single Element in a Sorted Array - You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. Return the single eleme
LeetCode 0540. Single Element in a Sorted Array有序数组中的单一元素【Medium】【Python】【二分】 Problem LeetCode You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. Find this single element that ...
Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once. Example 1: AI检测代码解析 Input: [1,1,2,3,3,4,4,8,8] Output: 2 1. 2. Example 2: AI检测代码解析 Input: [3...
[LeetCode] Single Number,Givenanarrayofintegers,everyelementappearstwiceexceptforone.Findthatsingleone.Note:Youralgorithmshouldhavealinearruntimecomplexity...
leetcode 136. Single Number Given an array of integers, every element appears twice except for one. ... LeetCode 260. Single Number III(只出现一次的数字 III) LeetCode 260. Single Number III(只出现一次的数字 III) LeetCode136 Single Number, LeetCode137 Single Number II, LeetCode260 Single...
LeetCode 0540. Single Element in a Sorted Array有序数组中的单一元素【Medium】【Python】【二分】 Problem LeetCode You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. Find this single element that ...
【leetcode】540. Single Element in a Sorted Array 题目如下: 解题思路:题目要求时间复杂度是O(logN),可以尝试使用二分查找法。首先数组是有序的,而且仅有一个元素出现一次,其余均为两次。我们可以先找到数组最中间的元素,记为mid。如果mid和mid-1以及mid+1都不相同,那么mid就是single number。如果mid和mid-...
如果nums[2 * mid] != nums[2 * mid + 1],那么说明单独的元素一定在前半段,否则就在后半段。这个题的官方题解写的很详细,我参考了方法三。 时间O(logn) 空间O(1) Java实现 1classSolution {2publicintsingleNonDuplicate(int[] nums) {3intstart = 0;4intend = nums.length / 2;5while(start ...
classSolution {public:intsingleNonDuplicate(vector<int>&nums) {intleft =0, right = nums.size() -1, n =nums.size();while(left <right) {intmid = left + (right - left) /2;if(nums[mid] == nums[mid +1]) {if((n -1- mid) %2==1) right =mid;elseleft = mid +1; ...