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.

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.

做这道题之前建议先看142. Linked List Cycle II，解题思路有些相通。这里有n+1个数，范围是1-n，也就是说这个数组可以看成是一个链表nums[n]。这里把n看做node，nums[n]看作next操作。链表的表头怎么选，这里只能选0，为什么呢？因为nums[0]可以保证向别的跳转，nums[0]不会等于0，如果选1做表头，万一nums[1] == 1，就会死循环。思路可以参考引文，代码如下：

public class Solution {    public int findDuplicate(int[] nums) {        int slow = nums[0];        int fast = nums[nums[0]];        while (slow != fast) {            slow = nums[slow];            fast = nums[nums[fast]];        }        fast = 0;        while (slow != fast) {            slow = nums[slow];            fast = nums[fast];        }        return slow;    }}