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.
题目标签：Hash Table
这道题让我们从两个list中找出相同的string并且它们的index sum是最小的，意思就是让我们找到他们两个都想吃的同一个饭店，并且这个饭店是他们排位里最靠前的。
Step 1: 首先我们建立一个HashMap， 遍历list1， 把饭店string作为key， 把index作为value保存进map。
Step 2: 建立一个HashSet, 遍历list2， 如果找到list2里的饭店string是在之前的map里的话，更新一下map的value = index1(之前的value） + index2（在list2里的）；并且设一个min，在记录最小的index sum，把最小的饭店string保存进set里面。
当找到一个相同的index sum = min的话，把这个饭店string加入set；当找到更小的min的时候，要把set清空，因为出现更小的index sum的饭店了，淘汰前面的饭店，重设min的值。
Step 3: 此时set里的饭店名就是最靠前的并且是两个list都有的，设置一个string array把答案copy进去return。
Solution:
Runtime beats 69.17%
完成日期：06/07/2017
1 public class Solution
2 {
3 public String[] findRestaurant(String[] list1, String[] list2)
4 {
5 HashMap<String, Integer> map = new HashMap<>();
6
7 // put list1‘s string as key, index as value.
8 for(int i=0; i<list1.length; i++)
9 map.put(list1[i], i);
10
11
12
13 HashSet<String> set = new HashSet<>();
14
15 int min = Integer.MAX_VALUE;
16
17 // iterate list2 to see any string is in map
18 for(int i=0; i<list2.length; i++)
19 {
20 if(map.get(list2[i]) != null) // if list2‘s string is in map
21 {
22 int j = map.get(list2[i]);
23 map.put(list2[i], i + j); // update map‘s value
24
25 if(i+j == min) // if find another same min value, add this string to set.
26 set.add(list2[i]);
27 else if(i+j < min) // if find another smaller index
28 {
29 set.clear(); // clear the set.
30 set.add(list2[i]); // add smaller index string into set.
31 min = i+j; // update the min;
32 }
33
34
35 }
36 }
37
38 String[] res = new String[set.size()];
39
40 int i=0;
41 for(String s : set)
42 {
43 res[i] = s;
44 i++;
45 }
46
47
48 return res;
49 }
50 }
参考资料：
http://blog.csdn.net/kangbin825/article/details/72794253