Find All Numbers Disappeared in an Array

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于 2018-05-20 04:40:00 发布

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CC 4.0 BY-SA版权

文章标签： runtime

原文链接：http://www.cnblogs.com/amazingzoe/p/9062361.html

本文介绍了一种在不使用额外空间且时间复杂度为O(n)的情况下找出数组中缺失元素的方法。通过将数组中的每个元素放置到其正确的位置上，然后扫描数组找到未在正确位置上的元素，从而确定缺失的数字。

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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]

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        // put all nums in correct place. nums[nums[i] - 1] = nums[i]
        for (int i = 0; i < nums.length;) {
            int newIndex = nums[i] - 1;
            if (i != newIndex && nums[i] != i + 1 && nums[i] != nums[newIndex]) {
                int temp = nums[newIndex];
                nums[newIndex] = nums[i];
                nums[i] = temp;
            } else {
                i++;
            }
        }
        
        // count the missing numbers
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                result.add(i + 1);
            } 
        }
        
        return result;
    }
}