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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]

我自己做的相当笨的方法，遍历套遍历，非常笨。代码如下：

public class Solution {
       public List
   
     findDisappearedNumbers(int[] nums) {        Arrays.sort(nums);        List
    
      list = new ArrayList
     
      ();        int flag = 0;        for(int i = 1; i <= nums.length; i++) {            while(true) {                if(i == nums[flag]) {                    if(flag < nums.length - 1) flag++;                    break;                }                if(i < nums[flag] || i > nums[nums.length - 1]) {                    list.add(i);                    break;                }                flag++;            }        }        return list;    }}
     
    
   

看到discuss里面的又跪了，简单解释下：

public List
   
     findDisappearedNumbers(int[] nums) {    List
    
      ret = new ArrayList
     
      ();    for(int i = 0; i < nums.length; i++) {        int val = Math.abs(nums[i]) - 1;        if(nums[val] > 0) {            nums[val] = -nums[val];        }    }    for(int i = 0; i < nums.length; i++) {        if(nums[i] > 0) {            ret.add(i+1);        }    }    return ret;}
     
    
   

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