【Leetcode】Find the Duplicate Number

题目链接：https://leetcode.com/problems/find-the-duplicate-number/

题目：

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：

不能修改数组所以不能排序，不能用额外的空间所以不能用HashMap，考虑鸽巢原理。数的范围在1~n之间，有n+1个数

猜测重复数为n/2，计算数组中大于n/2的次数count，如果重复的数在n/2~n之间的话，count应该是大于n-n/2的，搜索范围就减少为[n/2~n]了；否则搜索范围为[1~n/2]。

算法：

public int findDuplicate(int[] nums) {int n = nums.length-1,l=1,r=n;while(l<r){int m = (l+r)/2,count=0,t=0;for(int i=0;i<nums.length;i++){if(nums[i]>m)count++;if(nums[i]==m){t++;}}if(t>1)return m;if(count>(n-m)){ //当count大于m应该被大于的次数，则重复值肯定在右边l = m+1;}else{r = m-1;}}return l;}