LeetCode 287. Find the Duplicate Number

问题描述：

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than O(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

思路：

两种思路：

1、把nums排序，依次检查，如果位置 i 和 i+1 位置的元素相同，就是重复元素；

2、用二分查找，因为是把n个值放在n+1个位置中，所以每次计数有c个大于mid的值，如果c>mid 说明重复值在mid-high之间，返回说明重复值在low-mid之间，重复循环mid=（low+high）/2。

代码：

solution1：

int findDuplicate1(vector<int>& nums) {sort(nums.begin(), nums.end());for (int i = 0; i < nums.size() - 1; i++) {if (nums[i] == nums[i + 1])return nums[i];}}

solution2：

int findDuplicate2(vector<int>& nums) {int lo = 1, hi = nums.size() - 1;while (lo < hi) {int mid = (hi + lo) / 2;int c = 0;for (int i : nums)if (i <= mid)c += 1;if (c > mid)hi = mid;elselo = mid + 1;}return lo;}