題目：

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.

解答：

其實我第一個想法就是排序，如果前後相同就說明元素重複了。滿足上面的所有要求，然後AC了……

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = nums[0];
        for(int i = 0; i< nums.size() - 1; i++)
        {
            if(nums[i] == nums[i+1])    ans = nums[i];
        }
        return ans;
    }
};
 

1. 二分法：遍歷一遍陣列，大於中間數的個數如果多於 (n / 2) 則說明多出來的數在大於中間數的那部分中，同理也可能在小於中間數的那部分中。這樣每次只關注某一半的資料，搜尋一半而已；
2. 對映找環法：使用快慢兩個指標遍歷，非常巧妙