## 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:

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.

## 题目大意：

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

## 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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