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).

题目大意：

让我们找到坐标位置之和最小的相同的字符串

注意：

1. 列表长度范围[1, 1000]

2. 字符串长度[1, 30]

3. 下标范围[0, len - 1]

4. 列表内无重复

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，如果最小坐标和当前坐标和相同，那么将这个字符串加入结果result中，如果当前坐标和比最小坐标和小，那么更新为这个较小值，然后将结果result清空并加入这个字符串

注意：因为可能存在多个最小值，返回的是list，所以需要额外判断一下相等的情况

class Solution(object):
    def findRestaurant(self, list1, list2):
        """
        :type list1: List[str]
        :type list2: List[str]
        :rtype: List[str]
        """
        hashmap = {}
        result = []
        indexsum = len(list1) + len(list2)
        for i in xrange(len(list1)):
            hashmap[list1[i]] = i
        for i in xrange(len(list2)):
            if list2[i] in hashmap:
                if hashmap[list2[i]] + i < indexsum:
                    result = [list2[i]]
                    indexsum = hashmap[list2[i]] + i
                elif hashmap[list2[i]] + i == indexsum:
                    result.append(list2[i])
        return result