645.set_mismatch.md

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645.Set Mismatch

难度:Easy

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example 1:
Input: nums = [1,2,2,4]
Output: [2,3]
Note:
The given array size will in the range [2, 10000].
The given array's numbers won't have any order.

一次遍历统计次数，一次遍历找出结果。

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int>tmp(nums.size(),0);
        for(auto i:nums)
            tmp[i-1]++;
        vector<int> res={0,0};
        for(int i=0;i<nums.size();i++)
        {
            if(tmp[i]==2) res[0]=i+1;
            else if(tmp[i]==0) res[1]=i+1;
            if(res[0] && res[1]) break;
        }
        return res;
    }
};

执行用时 :32 ms, 在所有 C++ 提交中击败了98.14%的用户
内存消耗 :11 MB, 在所有 C++ 提交中击败了47.41%的用户