LeetCode 448 Find All Numbers Disappeared in an Array (思维推荐)

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本文介绍了一种在O(n)时间内找出数组中未出现的所有[1,n]范围内的元素的方法，无需额外空间，仅使用数组自身进行原地交换。通过实例[4,3,2,7,8,2,3,1]展示了算法步骤，最终输出[5,6]。该算法通过将数字i置于第i-1位置实现，最后检查不匹配的索引即可找到消失的数。

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]

题目链接：https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/

题目分析：有点像First Missing Positive，思想就是将数字i放到第i-1位上

6ms，时间击败92.45%，空间击败94.76%

class Solution {
    
    public void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
    
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> ans = new ArrayList<>();
        int pos = 0;
        while (pos < nums.length) {
            while (nums[pos] != pos + 1 && nums[pos] != nums[nums[pos] - 1]) {
                swap(nums, pos, nums[pos] - 1);
            }
            pos++;
        }
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                ans.add(i + 1);
            }
        }
        return ans;
    }
}

// 4 3 2 7 8 2 3 1
// 7 3 2 4 8 2 3 1
// 3 3 2 4 8 2 7 1
// 2 3 3 4 8 2 7 1
// 3 2 3 4 8 2 7 1
// 3 2 3 4 1 2 7 8
// 1 2 3 4 3 2 7 8