1.二分方法
因為數字是1~n，所以我們可以根據小於mid的數目來判斷重複的數字是在左邊還是右邊

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left = 1,right = nums.size()-1;
        while(left <= right){
            int mid = (left+right) >> 1;
            int cnt = 0;
            for(int x : nums)
                if(x <= mid)
                    cnt++;
            if(cnt > mid)
                right = mid-1;
            else
                left = mid+1;
        }
        return left;
    }
};
 

2.快慢指標
類似於142. Linked List Cycle II

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int fast = 0, slow = 0;
        while(true){
            fast = nums[nums[fast]];
            slow = nums[slow];
            if(fast == slow){
                fast = 0;
                while(fast != slow){
                    fast = nums[fast];
                    slow = nums[slow];
                }
                return slow;
            }
        }
        return 0;
    }
};