448. Find All Numbers Disappeared in an Array

 

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…)


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