[LeetCode] Find All Numbers Disappeared in an Array 找出数组中所有消失的数字,LeetCode All in One 题目讲解汇总(持续更新中...)

 

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]

 

这道题让我们找出数组中所有消失的数,跟之前那道Find All Duplicates in an Array极其类似,那道题让找出所有重复的数字,这道题让找不存在的数,这类问题的一个重要条件就是1 ≤ a[i] ≤ n (n = size of array),不然很难在O(1)空间和O(n)时间内完成。三种解法也跟之前题目的解法极其类似。首先来看第一种解法,这种解法的思路路是,对于每个数字nums[i],如果其对应的nums[nums[i] – 1]是正数,我们就赋值为其相反数,如果已经是负数了,就不变了,那么最后我们只要把留下的整数对应的位置加入结果res中即可,参见代码如下:

 

解法一:

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            int idx = abs(nums[i]) - 1;
            nums[idx] = (nums[idx] > 0) ? -nums[idx] : nums[idx];
        }
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] > 0) {
                res.push_back(i + 1);
            }
        }
        return res;
    }
};

 

第二种方法是将nums[i]置换到其对应的位置nums[nums[i]-1]上去,比如对于没有缺失项的正确的顺序应该是[1, 2, 3, 4, 5, 6, 7, 8],而我们现在却是[4,3,2,7,8,2,3,1],我们需要把数字移动到正确的位置上去,比如第一个4就应该和7先交换个位置,以此类推,最后得到的顺序应该是[1, 2, 3, 4, 3, 2, 7, 8],我们最后在对应位置检验,如果nums[i]和i+1不等,那么我们将i+1存入结果res中即可,参见代码如下:

 

解法二:

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] != nums[nums[i] - 1]) {
                swap(nums[i], nums[nums[i] - 1]);
                --i;
            }
        }
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] != i + 1) {
                res.push_back(i + 1);
            }
        }
        return res;
    }
};

 

下面这种方法是在nums[nums[i]-1]位置累加数组长度n,注意nums[i]-1有可能越界,所以我们需要对n取余,最后要找出缺失的数只需要看nums[i]的值是否小于等于n即可,最后遍历完nums[i]数组为[12, 19, 18, 15, 8, 2, 11, 9],我们发现有两个数字8和2小于等于n,那么就可以通过i+1来得到正确的结果5和6了,参见代码如下:

 

解法三:

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            nums[(nums[i] - 1) % n] += n;            
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] <= n) {
                res.push_back(i + 1);
            }
        }
        return res;
    }
};

 

类似题目:

Find All Duplicates in an Array

First Missing Positive

 

参考资料:

https://discuss.leetcode.com/topic/65944/c-solution-o-1-space

https://discuss.leetcode.com/topic/66063/5-line-java-easy-understanding

 

LeetCode All in One 题目讲解汇总(持续更新中…)

    原文作者:Grandyang
    原文地址: https://www.cnblogs.com/grandyang/p/6222149.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞