287. Find the Duplicate Number

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.

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 thanO(n2)
There is only one duplicate number in the array, but it could be repeated more than once

n + 1个位置，大小范围是1 - n，只能有一个duplicate的这些条件限定了只有1,2,2,3,4,5这种单一情况。所以用抽屉原理即可。

class Solution {
    public int findDuplicate(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 1;
        int end = nums.length - 1;
        int mid;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (countNum(nums, mid) <= mid) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (countNum(nums, start) > start) {
            return start;
        }
        return end;
    }

    private int countNum(int[] nums, int target) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] <= target) {
                count++;
            }
        }
        return count;
    }
}