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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Analysis:

if duplicated number is less than X, then there must be more than X numbers in the array less than X.

So, we use binary search from 1 to n, count the number of number <= mid, and determine whether is the duplicated number.

Solution:

public class Solution {    public int findDuplicate(int[] nums) {        if (nums.length == 0)            return 0;        int start = 1, end = nums.length - 1; // start is 1, end is n.        while (start <= end) {            int mid = start + (end - start) / 2;            // count the numbers no larger than mid.            int count = 0;            for (int num : nums)                if (num <= mid) {                    count++;                }            if (count > mid) {                end = mid - 1;            } else {                start = mid + 1;            }        }        return start;    }}

LeetCode-Find the Duplicate Number