LeetCode

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：给两个字符串数组，找出其中相同的字符串，返回在两个数组中索引相加最小的字符串，若有多个，则返回多个。

思路：用一个map把第一个数组全存起来，然后遍历第二个数组看有没有存在的字符串，记录最小索引和。再遍历一遍得出答案。时间复杂度O(n+m)，空间复杂度O(n)

class Solution {public:    vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {        unordered_map<string, int> vis;        vector<string> ans;        for (int i = 0; i < list1.size(); ++i) {            vis[list1[i]] = i;        }        int mmin = list1.size() + list2.size();        for (int i = 0; i < list2.size(); ++i) {            if (vis.find(list2[i]) == vis.end()) continue;            int x = i + vis[list2[i]];            mmin = min(mmin, x);        }        for (int i = 0; i < list2.size(); ++i) {            if (vis.find(list2[i]) == vis.end()) continue;            int x = i + vis[list2[i]];            if (x == mmin)                 ans.push_back(list2[i]);        }        return ans;    }};