[LeetCode] Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

分析一：利用数学公式：1+2+3+4+...+n = n(n+1)/2

　　要找的数就是和减去已有的数。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int count = nums.size();
        int sum = count * (count + 1) / 2;
        
        for (vector<int>::iterator iter = nums.begin(); iter != nums.end(); ++iter) {
            sum -= *iter;
        }
        
        return sum;
    }
};

分析二：利用位操作。

 int missingNumber(int nums[]) { //xor
    int res = nums.size();
    for(int i=0; i<nums.length; i++){
        res ^= i;
        res ^= nums[i];
    }
    return res;
}