内容简介:[LeetCode]Minimum Index Sum of Two Lists
题目描述:
LeetCode 599. Minimum Index Sum of Two Lists
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.
题目大意:
求两个字符串列表中索引之和最小的公共串
注意:
- 列表长度范围[1, 1000]
- 字符串长度[1, 30]
- 下标范围[0, len - 1]
- 列表内无重复
解题思路:
字典(Map)
Python代码:
class Solution(object):
def findRestaurant(self, list1, list2):
"""
:type list1: List[str]
:type list2: List[str]
:rtype: List[str]
"""
dict1 = {v : i for i, v in enumerate(list1)}
minSum = len(list1) + len(list2)
ans = []
for i, r in enumerate(list2):
if r not in dict1:
continue
currSum = i + dict1[r]
if currSum < minSum:
ans = [r]
minSum = currSum
elif currSum == minSum:
ans.append(r)
return ans
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