LeetCode 645. Set Mismatch

645. Set Mismatch

Description

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.

Solution

• 题意即一个数组中原本有1-n的数，但是现在有一个数重复并导致另一个数丢失，现在寻找重复的数和丢失的数。
• 我的做法是找一个桶（数组），来统计每一个数的个数，毕竟范围不超过20000，如果个数为0，即丢失的数字，个数为2，即重复的数字，代码如下

class Solution {public:    vector<int> findErrorNums(vector<int>& nums) {        vector<int> rnt;        int box[10010] = {0};        for (int i = 0;i < nums.size();i++) {            box[nums[i]]++;            if (box[nums[i]] == 2) rnt.push_back(nums[i]);        }        for (int i = 1;i <= nums.size();i++) {            if (box[i] == 0) {                rnt.push_back(i);                break;            }        }        return rnt;    }};

class Solution:    def findErrorNums(self, nums):        rnt = [];        box = [0] * 20000;        l = len(nums);        for i in nums:            box[i] += 1            if (box[i] == 2):                rnt.append(i);        for i in range(1,l + 1):            if box[i] == 0:                rnt.append(i);        return rnt;