LeetCode 540. Single Element in a Sorted Array (Medium)
2018-01-03 12:20
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题目描述:
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.Note: Your solution should run in O(log n) time and O(1) space.
Example1:
Input: [1,1,2,3,3,4,4,8,8] Output: 2
Example2:
Input: [3,3,7,7,10,11,11] Output: 10
题目大意:给一个排好序(从小到大)的数组,找出不是一对的那个元素(只出现1次),需要O(log n)时间复杂度以及O(1)空间复杂度。
思路:一看到O(log n)时间复杂度就马上联想到二分….然后,二分的前半的尾和后半的头有可能相同(一对),或者不同(不是一对),如果是一对的,前半加上这个后半的头,检查一下前半和后半的长度是否奇数,如果是,单独出现的数就在那一半,再继续二分直到某半只有一个元素即可。c++代码:
class Solution { public: int singleNonDuplicate(vector<int>& nums) { int l = 0; int r = nums.size() - 1; while (r != l) { int mid = (l + r) / 2; if (nums[mid] != nums[mid - 1]) { mid = mid - 1; } if ((mid + 1) % 2 == 0) { l = mid + 1; } else { r = mid; } } return nums[l]; } };
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