[LeetCode]645. Set Mismatch

其实还是那种利用【数组元素->数组下标】来【转换正负性】的问题。

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:

1. The given array size will in the range [2, 10000].
2. The given array's numbers won't have any order.

我的繁琐办法：

public class Solution {    public int[] findErrorNums(int[] nums) {        int[] res = new int[2];        if(nums==null || nums.length<2) return res;        int n=nums.length;                Map<Integer, Integer> map = new HashMap<>();        for(int i=0; i<nums.length; i++){            map.put(nums[i],map.getOrDefault(nums[i],0)+1);            map.put(i+1,map.getOrDefault(i+1,0)+1);        }        for(int key:map.keySet())            if(map.get(key)==3)                res[0]=key;                for(int key:map.keySet())            if(map.get(key)==1)                res[1]=key;                return res;    }}

方法二，利用下标：

public static int[] findErrorNums(int[] nums) {    int[] res = new int[2];    for (int i : nums) {        if (nums[Math.abs(i) - 1] < 0) res[0] = Math.abs(i);else nums[Math.abs(i) - 1] *= -1;    }    for (int i=0;i<nums.length;i++) {        if (nums[i] > 0) res[1] = i+1;    }    return res;}