287 Find the Duplicate Number

287. Find the Duplicate Number

1. Question

Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(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 thanO(n^2).

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

2. Implementation

(1) Binary Search

class Solution {
    public int findDuplicate(int[] nums) {
        int start = 0, end = nums.length - 1, mid = 0;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;

            int count = countNum(nums, mid);

            if (count <= mid) {
                start = mid + 1;
            }
            else {
                end = mid;
            }
        }

        return countNum(nums, start) > start ? start : end;
    }

    public int countNum(int[] nums, int target) {
        int count = 0;

        for (int num : nums) {
            if (num <= target) {
                ++count;
            }
        }
        return count;
    }
}

(2) Two Pointers

public class Solution {
    public int findDuplicate(int[] nums) {
        // Similar to Linked List Cycle II 
        // Use Two Pointers to find the begnning of loop
        // This is where the duplicate occur
        int fast = 0, slow = 0;

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

        slow = 0;

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

3. Time & Space Complexity

Binary Search: 时间复杂度O(nlogn), 空间复杂度O(1)

Two Pointers: 时间复杂度O(n), 空间复杂度O(1)