[Leetcode / hash table lists the minimum of two index] and (applying a hash map)

Problem Description:

Suppose Andy and Doris want to choose a restaurant for dinner, and they have expressed a favorite restaurant list, the name of each restaurant is represented by a string.

You need to help them with a minimum of index and find out their common favorite restaurant . If the answer is more than one output all the answers and do not consider the order. You can assume that there is always an answer.

Example 1:

Enter: 
[ "Shogun", "Tapioca Express", "Burger King", "KFC"] 
[ "Piatti", "at The Grill AT Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"] Output: [ "Shogun" ]
 explanation: their only common favorite restaurant is "Shogun".

Example 2:

Enter: 
[ "Shogun", "Tapioca Express", "Burger King", "KFC"] 
[ "KFC", "Shogun", "Burger King"] Output: [ "Shogun"]
 to explain: they co-favorite with minimal index and restaurant are "Shogun", which has the smallest index and 1 (0 + 1).

prompt:

1. Length two lists are [1, 1000] within.
2. Length of the string within the two lists will be in the range [1,30] of.
3. Index from 0 to 1 minus the length of the list.
4. Both lists are no duplicate elements.

The basic idea:

First establish repetitive elements and their indexing and mapping .

And then find the smallest index.

In order to return the index and the minimum index and the same for all the elements, then sweep again obtained .

AC Code:

class Solution {
public:
    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
      // 找到重复的字符串，记录他们重复的元素以及对应的索引和
      map<string, int> m;
      for (int i = 0; i < list1.size(); ++i) {
        auto it = find(list2.begin(), list2.end(), list1[i]);
        if (it != list2.end()) {
          m[list1[i]] = (it - list2.begin()) + i;
        } 
      }
      // 遍历hashmap，找到最小的索引和
      int min_index = INT_MAX;
      string min_string;
      for (auto i = m.begin(); i != m.end(); ++i) {     // hashmap的iterator使用pair的方式来访问key和value
         if (i->second < min_index) {
           min_index = i->second;
           min_string = i->first;
         }
      }
      // 返回结果集
      vector<string> res;
      for (auto i = m.begin(); i != m.end(); ++i) {
        if (i->second == min_index) {
          res.push_back(i->first);
        }
      }
      return res;
    }
};