leetcode 599. Minimum Index Sum of Two Lists 寻找index之和最小的公共元素

﹏ヽ暗。殇╰゛Y 2022-06-03 04:47 111阅读 0赞

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

本题题意有点绕，其实就是寻找公共元素，并且公共元素在该数组内的index之和是最小的，使用map做一个标记以供查询，然后遍历即可完成。

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    
    class Solution
    {
    public:
        vector<string> findRestaurant(vector<string>& list1, vector<string>& list2)
        {
            vector<string> res;
    
            map<string, int> m;
            for (int i = 0; i < list1.size(); ++i)
                m[list1[i]] = i;
    
            int minSum = INT_MAX;
            for (int i = 0; i < list2.size(); ++i)
            {
                if (m.count(list2[i]))
                {
                    int sum = i + m[list2[i]];
                    if (sum == minSum)
                        res.push_back(list2[i]);
                    else if (sum < minSum)
                    {
                        minSum = sum;
                        res = { list2[i] };
                    }
                }
            }
            return res;
        }
    };