LeetCode 645 : Set Mismatch（java）

原题

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.

思路

题意大致如下：set S的长度为n，无序存储了数字1-n，但由于data error，有一个数字出现了重复（同时意味着有一个数字缺失），找出重复数字和缺失的数字。
用一个map，第一遍循环，把set S过一遍，出现的数字用map标记，同时记录下重复的数字（已标记过的数字第二次读到为重复数字）。
第二遍循环，把map过一遍，查找缺失数字（没有被标记的数字）。
时间复杂度为o(n)。

代码

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int res[]=new int[2];
        //boolean类型默认初始化为false
        boolean map[]=new boolean[nums.length+1];
        for(int i=0;iif(map[nums[i]]==false)
                map[nums[i]]=true;
            else
                res[0]=nums[i];
        for(int i=1;i<(nums.length+1);i++)
            if(map[i]==false){
                res[1]=i;
                break;
            }
        return res;
    }
}