leetcode解题之448. Find All Numbers Disappeared in an Array Java版 (找出缺少的数字）

448. Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

不使用额外的空间找出数组中缺少的数字，有几个缺少的就有几个重复的
import java.util.*;

public class Solution {
// 注意有几个缺少的就有几个重复的。总数量不变，标志位解决
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> ret = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
// nums[i]从1到n而下标从0开始，要减去1
// temp是从[0,n-1]中少两个,但是有重复
int temp = Math.abs(nums[i]) - 1;
if (nums[temp] > 0)
nums[temp] = -nums[temp];
}
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0)
ret.add(i + 1);
}
return ret;
}
}

import java.util.*;
public class Solution {
// 使用set存储出现的元素
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> ret = new ArrayList<>();
// 不能使用list，有重复元素，容易超时
Set<Integer> temp = new HashSet<>();
for (int i : nums) {
temp.add(i);
}
for (int i = 1; i <= nums.length; i++) {
if (!temp.contains(i))
ret.add(i);
}
return ret;
}

}