LC 599. Minimum Index Sum of Two Lists

题目描述

Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.

You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.

Example 1:

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".

Example 2:

Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).

参考答案

 class Solution {
 public:
     vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {

         vector<string>  res;
         unordered_map<string, int> map;
         for(size_t i =  ; i< list1.size(); i++)
             map[list1[i]] = i ; // list1 = key ; i = value ;

         size_t  min = INT_MAX;

         for(size_t  i =  ; i<list2.size();i ++){
             auto iter = map.find(list2[i]);
             if(iter != map.end()){
                 if(iter->second + i< min ){
                     min = map[list2[i]] + i;
                     res.assign(,list2[i]); // 直接把值赋给第一个，不需要clear然后重新赋值

                 }else if(iter->second + i  == min){
                     res.push_back(list2[i]);
                 }

             }
         }
         return res;
     }
 };

分析备注

1.  for 循环中，int 替换为 size_t 可以加速。

2.  vector.assign(1,value) 可以实现 清除+赋值 两步操作。

3.  使用 iter -> first 和 iter -> second 可以更快。