【LeetCode从零单刷】Find the Duplicate Number

题目：

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：

其实我第一个想法就是排序，如果前后相同就说明元素重复了。满足上面的所有要求，然后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：遍历一遍数组，大于中间数的个数如果多于 (n / 2) 则说明多出来的数在大于中间数的那部分中，同理也可能在小于中间数的那部分中。这样每次只关注某一半的数据，搜索一半而已；
2. 映射找环法：使用快慢两个指针遍历，非常巧妙

【LeetCode从零单刷】Find the Duplicate Number

原文：http://blog.csdn.net/ironyoung/article/details/49018559