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.
  
  题目的大意是找到两个数组中相同元素的index和最小的元素。解法是将数组a的元素放入map中，值为key、index为value， 当数组b也出现同样的元素的时候，两个元素的index的和与当前的最小值判断，大就舍弃，小就更新最小值，并放入一个新的列表。

public static String[] findRestaurant(String[] list1, String[] list2) {

        int min = Integer.MAX_VALUE;

        List<String>  result = new ArrayList<>();

        HashMap<String,Integer> map = new HashMap<>();

        for(int i = 0; i < list1.length; i++)

            map.put(list1[i],i);

        for(int i = 0; i < list2.length; i++)

        {

            if(map.containsKey(list2[i]))

            {

                int index = map.get(list2[i]);

                if(i + index == min)

                {

                    result.add(list2[i]);

                }

                else if(i + index < min)

                {

                    result.clear();

                    result.add(list2[i]);

                }

            }

        }

        return result.toArray(new String[result.size()]); //这里将list转为数值

    }

看了leetcode上面的Discuss其他用户的写法，上面的代码可以做一个小小的优化，

    public static String[] findRestaurant(String[] list1, String[] list2) {

        int min = Integer.MAX_VALUE;

        List<String>  result = new ArrayList<>();

        HashMap<String,Integer> map = new HashMap<>();

        for(int i = 0; i < list1.length; i++)

            map.put(list1[i],i);

        for(int i = 0; i < list2.length; i++)

        {

            Integer index = map.get(list2[i]); /////

            if(index != null && i + index <= min)

            {

                if(i + index < min)

                {

                    result.clear(); //存在更小的值，就将列表清空

                    min = i + index;

                }

                result.add(list2[i]);

            }

        }

        return result.toArray(new String[result.size()]); //这里将list转为数值

    }

这是一个常规的题目，主要是需要注意的有以下几点

• 在选择较小的值的时候，往往为min赋值一个最大的值，注意写法
• 判断map是否包含某个key使用containsKey返回的为布尔类型，可以灵活使用Integer
• 清空列表使用clear，注意列表转数组的用法，数组转列表使用Arrays.asList(a)