599. Minimum Index Sum of Two Lists

题目：

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.

思路：

本题思路很简单，就便利比较即可，使用两层for循环

代码：

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
        int min = 2000,count;
        vector<string> str;
        for(int i=0;i<list1.size();i++)
        {
            for(int j =0;j<list2.size();j++)
            {
                if(list1[i]==list2[j])
                {
                    count=i+j;
                    if(min>count)
                    {
                        str.clear();
                        str.push_back(list1[i]);
                        min = count;
                    }
                    else if(min==count)
                    {
                        str.push_back(list1[i]);
                    }
                    break;
                }
            }
        }
        return str;
    }
};