287. Find-the-Duplicate-Number

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

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.

Method One

class Solution {
    public int findDuplicate(int[] nums) {
        // 这个特殊在 n + 1的长度里面的整数是从 1 - n 的，因此 index 就可以用来做标记。
        // 从 0 出发，就可以找到，这个和那个 LinkedList 一样
        int slow = nums[0];
        int fast= nums[nums[0]];

        while(slow != fast) {
            slow = nums[slow];
            fast = nums[nums[fast]];
        }

        slow = 0;
        while(slow != fast) {
            slow = nums[slow];
            fast = nums[fast];
        }

        return slow;
    }
}
​