[LeetCode] Set Mismatch

标签：solution   通过   ica   使用   original   one   div   vector   form   

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:

1. The given array size will in the range [2, 10000].
2. The given array‘s numbers won‘t have any order.

给定一个含有元素1~n的数组，由于数据错误，导致这个数组中有一个数字(num)发生了重复替换了另外一个数字(mis)，让我们找出这个重复的数字和被替换掉的数字。使用map找出重复的元素，也就是数组中出现2次的数字。使用set找出被替换掉的数字。set存放数组中的不相同元素，通过遍历1~n来找出那个被替换掉的数字。

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int num = 0, mis = 0;
        if (nums.size() == 0)
            return {};
        unordered_map<int, int> m;
        unordered_set<int> s(nums.begin(), nums.end());
        for (int num : nums)
            m[num]++;
        for (auto it = m.begin(); it != m.end(); it++) 
            if (it->second == 2) {
                num = it->first;
                break;
            }
        for (int i = 1; i <= nums.size(); i++)
            if (s.find(i) == s.end())
                mis = i;
        return {num, mis};
    }
};
// 82 ms

[LeetCode] Set Mismatch

标签：solution   通过   ica   使用   original   one   div   vector   form   

原文地址：http://www.cnblogs.com/immjc/p/7225321.html