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.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

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

      public int findDuplicate(int[] nums) {
           //二分搜素法
          //我们在区别[1, n]中搜索，首先求出中点mid，然后遍历整个数组，统计所有小于等于mid的数的个数，如果个数大于mid，则说明重复值在[mid+1, n]之间，反之，重复值应在[1, mid-1]之间，然后依次类推，知道搜索完成，此时的low就是我们要求的重复值
           int low = 1, high = nums.length - 1;
           while (low <= high) {
               int mid = low + (high - low)  / 2;
               int cnt = 0;
               for (int a : nums) {
                   if (a <= mid) ++cnt;
               }
               if (cnt <= mid) low = mid + 1;
               else high = mid - 1;
           }
           return low;
       }
   
   
           
       
       
       
       
       
       
       //快慢指针问题（思想比较好，也是平时面试中的经典变形题）
       //参见  http://bookshadow.com/weblog/2015/09/28/leetcode-find-duplicate-number/
   //     if (nums.length > 1)
   // 	{
   // 		int slow = nums[0];
   // 		int fast = nums[nums[0]];
   // 		while (slow != fast)
   // 		{
   // 			slow = nums[slow];
   // 			fast = nums[nums[fast]];
   // 		}
   
   // 		fast = 0;
   // 		while (fast != slow)
   // 		{
   // 			fast = nums[fast];
   // 			slow = nums[slow];
   // 		}
   // 		return slow;
   // 	}
   // 	return -1;