Intersection of Two Arrays II
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.
- 思路:先将两个数组排序,这样被比较数组遍历一次即可。nums1中的元素,从前到后在nums2中查找
- 分三种情况:
- 1. 如果nums2的元素大于nums1的当前元素,说明nums2中不存在此元素。
- 2. 如果nums2的元素等于nums1的当前元素,那么该元素为两个数组的交集,两个指针都前进。
- 3. 如果nums2的元素小于nums的当前元素,nums2的指针继续向后找。
- 具体实现如下:
-
class Solution { public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { vector<int> result; sort(nums1.begin(),nums1.end()); sort(nums2.begin(),nums2.end()); int ind1=0; int ind2 = 0; while(ind1<nums1.size() && ind2<nums2.size()){ if(nums1[ind1]<nums2[ind2]){ ind1++; }else if(nums1[ind1]==nums2[ind2]){ result.push_back(nums1[ind1]); ind1++; ind2++; }else{ ind2++; } } return result; } };