leetcode448-Find All Numbers Disappeared in an Array

•  问题描述

  • 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.

• 示例：

  Example:

  Input:
  [4,3,2,7,8,2,3,1]
  
  Output:
  [5,6]

• 测试用例：

• 输入数组为空：[]
• 输入数组长度为1：[1]
• [4,3,2,7,8,2,3,1]
• [1,2,3]

• 思路分析：

  • 第一次：因为数组内所有元素都在[1,n]之间，其中n为数组长度。故而假设数组没有重复的数组，那么数组排序之后就会由[1,2,3,.....,n]此时a[i] = i+1,因此我们可任意遍历两次数组， 第一次遍历时，将数组元素放在它排序后的位置，使a[i] = i + 1, 之后再次遍历时一次为判断条件， 当数组内元素不满足该条件时，i+1就是所缺失的元素。此方法不需额外的存储空间，时间复杂度为遍历数组的时间O(n),满足要求

    class Solution {
        public List<Integer> findDisappearedNumbers(int[] nums) {
            List<Integer> list = new ArrayList<>();
            if(nums.length <= 1) return list;
            for(int i = 0; i < nums.length; i++){
                while(nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]){
                    int temp = nums[i];
                    nums[i] = nums[nums[i] - 1];
                    nums[temp - 1] = temp;
                }
            }
    
            for(int i = 0; i < nums.length; i++){
                if(nums[i] != i + 1){
                    list.add(i + 1);
                }
            }
            return list;
        }
    }