350. Intersection of Two Arrays II
Posted 积少成多
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了350. Intersection of Two Arrays II相关的知识,希望对你有一定的参考价值。
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1]
, nums2 = [2, 2]
, return [2, 2]
.
Note:
- Each element in the result should appear as many times as it shows in both arrays.
- The result can be in any order.
Follow up:
- What if the given array is already sorted? How would you optimize your algorithm?
- What if nums1‘s size is small compared to nums2‘s size? Which algorithm is better?
- What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
===================
返回交叉元素,
注意:结果中每个元素应该出现和两个数组相同的次数,结果的顺序任意.
思路:排序+遍历
==
class Solution { public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { vector<int> re; sort(nums1.begin(),nums1.end()); sort(nums2.begin(),nums2.end()); vector<int>::iterator it1 = nums1.begin(); vector<int>::iterator it2 = nums2.begin(); for(;it1!=nums1.end() && it2!=nums2.end();){ if(*it1<*it2) it1++; else if(*it1 > *it2) it2++; else{ re.push_back(*it1); it1++; it2++; } } return re; } };
以上是关于350. Intersection of Two Arrays II的主要内容,如果未能解决你的问题,请参考以下文章
350. Intersection of Two Arrays II
Leetcode 350. Intersection of Two Arrays II
350. Intersection of Two Arrays II
350. Intersection of Two Arrays II