LeetCode 599. Minimum Index Sum of Two Lists （从两个lists里找到相同的并且位置总和最靠前的）

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。

　　　

Java Solution:

Runtime beats 69.17%

完成日期：06/07/2017

 public class Solution
 {
     public String[] findRestaurant(String[] list1, String[] list2)
     {
         HashMap<String, Integer> map = new HashMap<>();

         // put list1's string as key, index as value.
         for(int i=0; i<list1.length; i++)
             map.put(list1[i], i);

         HashSet<String> set = new HashSet<>();

         int min = Integer.MAX_VALUE;

         // iterate list2 to see any string is in map
         for(int i=0; i<list2.length; i++)
         {
             if(map.get(list2[i]) != null)    // if list2's string is in map
             {
                 int j = map.get(list2[i]);
                 map.put(list2[i], i + j); // update map's value

                 if(i+j == min)    // if find another same min value, add this string to set.
                     set.add(list2[i]);
                 else if(i+j < min)    // if find another smaller index
                 {
                     set.clear();    // clear the set.
                     set.add(list2[i]);    // add smaller index string into set.
                     min = i+j;    // update the min;
                 }

             }
         }

         String[] res = new String[set.size()];

         int i=0;
         for(String s : set)
         {
             res[i] = s;
             i++;
         }

         return res;
     }
 }

参考资料：

http://blog.csdn.net/kangbin825/article/details/72794253

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