287. Find the Duplicate Number (Medium)

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(n^2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

Solution 1: Binary Search

Time Complexity: $$O(nlogn)$$

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int low = 1, high = nums.size() - 1;
        while (low < high) {
            int mid = low + (high - low) * 0.5;
            int cnt = 0;
            for (auto a : nums) {
                if (a <= mid) ++cnt;
            }
            if (cnt <= mid) low = mid + 1;
            else high = mid;
        }
        return low;
    }
};

Solution 2: Two Pointers

经过热心网友waruzhi的留言提醒还有一种O(n)的解法，并给了参考帖子，发现真是一种不错的解法，其核心思想快慢指针在之前的题目Linked List Cycle II中就有应用，这里应用的更加巧妙一些，由于题目限定了区间[1,n]，所以可以巧妙的利用坐标和数值之间相互转换，而由于重复数字的存在，那么一定会形成环，我们用快慢指针可以找到环并确定环的起始位置，确实是太巧妙了！

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        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;

    }
};