lintcode(633)Find the Duplicate Number

描述：

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.

 注意事项

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(n^2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

样例：

Given nums = [5,5,4,3,2,1] return 5
Given nums = [5,4,4,3,2,1] return 4

思路：

建立i到f(A[i])的映射，会找到i!=j,而f(i)=f(j),进行环检测，查找包含重复值的循环，循环的入口即为重复值

详细证明参考：http://bookshadow.com/weblog/2015/09/28/leetcode-find-duplicate-number/

解题思路参考了此链接的博客内容

public class Solution {    /**     * @param nums an array containing n + 1 integers which is between 1 and n     * @return the duplicate one     */    public int findDuplicate(int[] nums) {        // Write your code         int slow = 0;        int fast = 0;        while(true){            slow = nums[slow];            fast = nums[nums[fast]];                        if(slow == fast){                break;            }        }                int result = 0;        while(true){            slow = nums[slow];            result = nums[result];                        if(slow == result){                return result;            }        }    }}