LeetCode.599 Minimum Index Sum of Two Lists （经典：哈希表对字符串去重和List与数组间的转换）

题目：

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.

分析：

class Solution {
public String[] findRestaurant(String[] list1, String[] list2) {
//给定两个菜单的字符串数组，找出其中相同的菜单下标和最小的下标的菜单名，如果存在下标和相同，则全部输出
//注意：菜单项最多1000，每个菜单的名字最多30个字符，下标从0开始计算。每个菜单中不存在重复的菜单名。
//思路：先求出相同项，然后下标和相同的存入相同的List数组中
List<String> list=new ArrayList<>();;

//参考队列
HashMap<String,Integer> hm=new HashMap<String,Integer>();

for(int i=0;i<list1.length;i++){
hm.put(list1[i],i);
}
int min=Integer.MAX_VALUE;
for(int i=0;i<list2.length;i++){
Integer j=hm.get(list2[i]);
if(j!=null&&i+j<=min){
//清楚list中的数据
if(i+j<min){
//存在更小的，清空之前的
list.clear();
min=i+j;
}
//添加元素，相同下标和
list.add(list2[i]);
}

}

//输出结果
return list.toArray(new String[list.size()]);
}
}