LeetCode:Find the Duplicate Number

Find the Duplicate Number

Total Accepted: 26713  Total Submissions: 68841  Difficulty: Hard

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.

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思路：

（1）直接看个例子

    index: 0 1 2 3 4 5 6 7 8

    nums: 2 1 4 7 5 6 4 3 8

（2）从index = 0出发，可以看到有这样一个循环：

           index: 0 -> 2 -> 6 -> 4 -> 6 ->...

nums[index]: 2 -> 4 -> 5 -> 6 -> 4->...

（3）可以看到从[4->5->6->4]是一个循环，正好4就是我们要求的duplicate number。

是不是有点像：Linked List Cycle II 中找环的入口点。

c++ code:

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        
        int size = nums.size();
        
        if(size < 2) 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;
    }
};