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.
這道題給了我們兩個字符串數組,讓我們找到坐標位置之和最小的相同的字符串。那么對於這種數組項和其坐標之間關系的題,最先考慮到的就是要建立數據和其位置坐標之間的映射。我們建立list1的值和坐標的之間的映射,然后遍歷list2,如果當前遍歷到的字符串在list1中也出現了,那么我們計算兩個的坐標之和,如果跟我們維護的最小坐標和mn相同,那么將這個字符串加入結果res中,如果比mn小,那么mn更新為這個較小值,然后將結果res清空並加入這個字符串,參見代碼如下:
class Solution { public: vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) { vector<string> res; unordered_map<string, int> m; int mn = INT_MAX, n1 = list1.size(), n2 = list2.size(); for (int i = 0; i < n1; ++i) m[list1[i]] = i; for (int i = 0; i < n2; ++i) { if (m.count(list2[i])) { int sum = i + m[list2[i]]; if (sum == mn) res.push_back(list2[i]); else if (sum < mn) { mn = sum; res = {list2[i]}; } } } return res; } };
類似題目:
Intersection of Two Linked Lists
參考資料:
https://discuss.leetcode.com/topic/90534/java-o-n-m-time-o-n-space