注意事項 Each element in the result must be unique. The result can be in any order. 樣例 nums1 = [1, 2, 2, 1], nums2 = [2, 2], 返回 [2].
熟悉hash解法嘗試 負值失敗 嘗試OFFSET?
#include <stdio.h>#include <vector>using namespace std;class Solution {public: /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { // Write your code here int i=0,j=0,k=0; int m=nums1.size(); int n=nums2.size(); int *Hash = new int[m+n]; vector<int> nums3; for(int p=0; p<=m+n-1;p++) Hash[p]=0; while(i <= m-1){ int temp=0; temp = nums1[i]; if(Hash[temp] == 0) Hash[temp]++; i++; } while(j <= n-1){ int temp=0; temp = nums2[j]; if(Hash[temp] == 1) Hash[temp]++; j++; } while(k <= m+n-1){ if(Hash[k] == 2) nums3.push_back(k); k++; } delete[] Hash; return nums3; }};第二種解法 常規 先排序后去重
#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>using namespace std;class Solution {public: /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { // Write your code here int i=0,j=0; int m=nums1.size(); int n=nums2.size(); vector<int> nums3; if(m==0 || n==0) return nums3; sort(nums1.begin(),nums1.end()); sort(nums2.begin(),nums2.end()); int temp; while(i <= m-1 && j <= n-1 ){ if(nums1[i] == nums2[j]){ if(temp != nums1[i]){ nums3.push_back(nums1[i]); temp = nums1[i]; } i++; j++; }else if(nums1[i] < nums2[j]){ i++; }else j++; } return nums3; }};新聞熱點
疑難解答
