leetCode-Find the Duplicate Number

Description:
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:

    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.

Solution:

//nums中元素对应的值作为下标,fast比slow快一步，如果fast=slow,说明fast和slow在一个环中.而重复元素正是这个环的入口,那么初始化一个fast下标为0,如果用同样的step再次遇到slow，即为环的入口，也就是重复元素class Solution {    public int findDuplicate(int[] nums) {        if (nums.length == 0) return -1;        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;    }}