LeetCode //C - 540. Single Element in a Sorted Array

540. 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 element that appears only once.

Your solution must run in O(log n) time and O(1) space.

Example 1:

Input: nums = [1,1,2,3,3,4,4,8,8]
Output: 2

Example 2:

Input: nums = [3,3,7,7,10,11,11]
Output: 10

Constraints:
  • 1 < = n u m s . l e n g t h < = 1 0 5 1 <= nums.length <= 10^5 1<=nums.length<=105
  • 0 < = n u m s [ i ] < = 1 0 5 0 <= nums[i] <= 10^5 0<=nums[i]<=105

From: LeetCode

Link: 540. Single Element in a Sorted Array


Solution:

Ideas:

1. Pairs Observation: In a perfectly paired array where each element appears exactly twice, if you pick any even index (0, 2, 4,...), the next index should have the same value if there's no single element disrupting the pairing.

2. Detecting the Single Element Influence: When the single element is introduced, it disrupts these pairings. Specifically, the first occurrence of every element in a disrupted array will start appearing at an even index and end at an odd index (before the single element appears).

3. Binary Search Strategy:

  • Check the middle index of the array.
  • If the middle index is even, then check the next element. If they are the same, it means the single element is ahead; otherwise, it's behind.
  • If the middle index is odd, then check the previous element. If they are the same, it means the single element is ahead; otherwise, it's behind.
  • Adjust the search boundaries accordingly until you find the single element.
Code:
c 复制代码
int singleNonDuplicate(int* nums, int numsSize) {
    int low = 0, high = numsSize - 1;
    while (low < high) {
        int mid = low + (high - low) / 2;
        // Ensure mid is even. If mid is odd, decrease by 1 to make it even.
        if (mid % 2 == 1) mid--;

        // Check pairs: nums[mid] should be the same as nums[mid + 1]
        if (nums[mid] == nums[mid + 1]) {
            // If they are the same, move to the right half
            low = mid + 2;
        } else {
            // If not the same, move to the left half
            high = mid;
        }
    }
    // Low should point to the single element
    return nums[low];
}
相关推荐
.Cnn几秒前
用邻接矩阵实现图的深度优先遍历
c语言·数据结构·算法·深度优先·图论
2401_858286116 分钟前
101.【C语言】数据结构之二叉树的堆实现(顺序结构) 下
c语言·开发语言·数据结构·算法·
Beau_Will12 分钟前
数据结构-树状数组专题(1)
数据结构·c++·算法
迷迭所归处16 分钟前
动态规划 —— 子数组系列-单词拆分
算法·动态规划
爱吃烤鸡翅的酸菜鱼17 分钟前
Java算法OJ(8)随机选择算法
java·数据结构·算法·排序算法
寻找码源1 小时前
【头歌实训:利用kmp算法求子串在主串中不重叠出现的次数】
c语言·数据结构·算法·字符串·kmp
Matlab精灵1 小时前
Matlab科研绘图:自定义内置多款配色函数
算法·matlab
诚丞成1 小时前
滑动窗口篇——如行云流水般的高效解法与智能之道(1)
算法
带多刺的玫瑰3 小时前
Leecode刷题C语言之统计不是特殊数字的数字数量
java·c语言·算法
爱敲代码的憨仔3 小时前
《线性代数的本质》
线性代数·算法·决策树