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.
这道题给了我们两个字符串数组，让我们找到坐标位置之和最小的相同的字符串。那么对于这种数组项和其坐标之间关系的题，最先考虑到的就是要建立数据和其位置坐标之间的映射。我们建立list1的值和坐标的之间的映射，然后遍历list2，如果当前遍历到的字符串在list1中也出现了，那么我们计算两个的坐标之和，如果跟我们维护的最小坐标和mn相同，那么将这个字符串加入结果res中，如果比mn小，那么mn更新为这个较小值，然后将结果res清空并加入这个字符串，参见代码如下：
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;
}
};
类似题目：
Intersection of Two Linked Lists
参考资料：
https://discuss.leetcode.com/topic/90534/java-o-n-m-time-o-n-space
LeetCode All in One 题目讲解汇总(持续更新中…)
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