【LeetCode】599、两个列表的最小索引总和

599、Minimum Index Sum of Two Lists 两个列表的最小索引总和

难度：简单

题目描述

• 英文：

  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.

• 中文：

  假设Andy和Doris想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。

  你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设总是存在一个答案。

示例

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).

提示

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.

代码提交

Python2，用时128ms，内存11.1M

class Solution(object):
    def findRestaurant(self, list1, list2):
        """
        :type list1: List[str]
        :type list2: List[str]
        :rtype: List[str]
        """
        dict1 = {}
        for index, name in enumerate(list1):
            dict1[name] = index
        
        templist = []
        tempsum = 3000
        for index, name in enumerate(list2):
            if name in dict1:
                cursum = index+dict1[name]
                if cursum < tempsum:
                    templist = [name]
                    tempsum = cursum
                elif cursum == tempsum:
                    templist.append(name)
        return templist

Tips

Python查找dict的key时，应使用if key in dict，而不是if key in dict.keys()。

前者是在dict中查找，dict对象的存储结构是hash表，最优情况下查询复杂度为O(1)；

后者等于是在list中查找，list对象的存储结构是线性表，查询复杂度为O(n)。