287 Find the Duplicate Number . *****要求比较奇葩 O(1)space，no modify array

Good Question

Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(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

You must not modify the array (assume the array is read only).

You must use only constant, O(1) extra space.

Your runtime complexity should be less than O(n2).

There is only one duplicate number in the array, but it could be repeated more than once.'

class Solution {
    public int findDuplicate(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] == nums[j]) {
                    return nums[i];
                }
            }
        }
        return 0;
    }
}

Binary Search 不可思议！！

time O(nlogn) space o(1)

class Solution {
    public int findDuplicate(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int start = 0;  
        int end = nums.length - 1;
        // [1,3,4,2,2]
        // [0,1,2,3,4] 
        while (start <= end) { // logn
            int mid = start + (end - start) / 2; //mid = 2 mid = 0// mid = 1
            int count = 0;
            for (int i = 0; i < nums.length; i++) {  // O(n)
                if(nums[i] <= mid) {
                    count++;//count = 3
                }
            }
            if (count > mid) {
                end = mid - 1;// end = 1, start = 0
            } else { //
                start = mid + 1;// start = 1// start = 2
            }
        }
        return start;
    }
}