Find the Duplicate Number--循环链表

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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.

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

Find the Duplicate Number--循环链表

标签：space   int   ica   fast   ast   bsp   out   The   odi   

原文地址：https://www.cnblogs.com/liudebo/p/9339061.html