题目来源【Leetcode】
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]
这道题比较简单,用一个布尔数组来判断就好,代码如下:
class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
int s = nums.size();
bool exit[s+1] = {false};
for(int i = 0; i < s; i ++){
exit[nums[i]] = true;
}
vector<int>disapp;
for(int i = 1; i <= s; i++){
if(exit[i] == false)
disapp.push_back(i);
}
return disapp;
}
};

本文介绍了一个LeetCode上的算法题目——寻找消失的数。题目要求在给定数组中找到所有未出现的1到n之间的整数,并且要在O(n)的时间复杂度内完成,不能使用额外的空间。文章提供了一种解决方案,利用布尔数组标记已出现的数字,最后返回未被标记的位置对应的下标。
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