LeetCode - 599 - Minimum Index Sum of Two Lists
2017-07-20 16:42
288 查看
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:
Example 2:
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.
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;
}
};
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.
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;
}
};
相关文章推荐
- Leetcode 599 Minimum Index Sum of Two Lists
- LeetCode 599 Minimum Index Sum of Two Lists
- 599 Minimum Index Sum of Two Lists
- HashTable-599-Minimum Index Sum of Two Lists
- LeetCode Minimum Index Sum of Two Lists
- [LeetCode] Minimum Index Sum of Two Lists 两个表单的最小坐标和
- leetcode[Minimum Index Sum of Two Lists]//待整理多种解法
- Minimum Index Sum of Two Lists
- Minimum Index Sum of Two Lists-python
- Minimum Index Sum of Two Lists问题及解法
- LeetCode.599 Minimum Index Sum of Two Lists (经典:哈希表对字符串去重和List与数组间的转换)
- LC-Minimum Index Sum of Two Lists
- 599.minimum-index-sum-of-two-lists
- Leetcode_371_Sum of Two Integers
- [LeetCode] Intersection of Two Linked Lists
- 【LeetCode】160 - Intersection of Two Linked Lists
- LeetCode(160) Intersection of Two Linked Lists
- Leetcode Intersection of Two Linked Lists
- [Leetcode] Sum of Two Integers
- LeetCode编程练习 - Intersection of Two Linked Lists学习心得