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 are only one duplicate number, find the duplicate one.

Note:

1. You must the Modify the array (assume the array was read only).
2. You must use only constant, O(1) Extra space.
3. Your runtime complexity should is less than O(n2) .
4. There is a duplicate number in the array, but it could was repeated more than once.

Credits:
Special thanks to @jianchao. Li.fighter for adding the problem and creating all test cases.

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Hide TagsBinary Search Array of pointersHide Similar Problems(H) First Missing Positive (m) single number (m) Linked List Cycle II (m) Missing number

Ideas:

(1) Take a direct look at an example

Index:0 1 2 3 4 5 6 7 8

Nums:2 1 4 7 5 6 4 3 8

(2) starting from index = 0, you can see that there is a loop:

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

Nums[index]: 2, 4, 5, 6, 4-> .....

(3) can be seen from [4->5->6->4] is a loop, exactly 4 is the duplicate number we requested.

Is it a bit like: The entry point of the Linked list Cycle II in the Find ring.

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;}    ;

Leetcode:find the Duplicate number