两个列表的最小索引总和

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).

 

Note:

1. The length of both lists will be in the range of [1, 1000].
2. The length of strings in both lists will be in the range of [1, 30].
3. The index is starting from 0 to the list length minus 1.
4. No duplicates in both lists.

 

题意：返回两个string数组中出现的，相同的且下标和最小的string,如果有多个这样的string，要把它们全部输出

思路：将数组1放入哈希表中，然后遍历数组2，如果有相同的就计算索引和，如果比之前的索引和小，那么更新最小索引和，并且清空结果数组，如果和之前的相等，就将其插入到结果数组中。

//将list1放入map中，遍历list2,如果有相同的就计算索引和
//if<之前的，更新索引和，清空ans，并将该字符串插入ans
//if==之前的，就插入ans

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        unordered_map<string,int> m;
        vector<string> ans;
        int sum=list1.size()+list2.size();
        for(int i=0;i<list1.size();i++){
            m[list1[i]]=i;
        }
        for(int i=0;i<list2.size();i++){
            if(m.count(list2[i])){
                int temp=i+m[list2[i]];
                if(temp<sum){
                    sum=temp;
                    ans.clear();
                    ans.push_back(list2[i]);
                }   
                else if(temp==sum){
                    ans.push_back(list2[i]);
                }
            }
        }
        return ans;   
    }
};

 

 

转载于:https://www.cnblogs.com/Bipolard/p/9994479.html