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.

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:


Example 2:

Input: ["Shogun", "Tapioca Express", "Burger King", "KFC"] ["KFC", "Shogun", "Burger King"] Output: ["Shogun"] Explanation:


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：
有两个字符串数组，找到他们当中相同的字符串，并求两个字符串的索引（数组的下标） 之和。把那些索引之和最小的字符串输出来。
注意，两个数组内没有相互重复的字符串。

思路：
看了Discuss中的方法。使用HashMap空间换时间，能够达到O(n+m)的复杂度。

首先，使用一个HashMap保存第一个字符串数组，key对应字符串，value对应索引。

然后，遍历第二个字符串，如果在HashMap中存在，则求索引之和，并记录到HashMap2中，key对应字符串，value对应索引之和。同时记录当前最小的索引值，记为min

最后，遍历HashMap2，把value等于min的全部输出

Java代码：

public class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {

        HashMap<String, Integer> map1 = new HashMap<>();
        HashMap<String, Integer> map2 = new HashMap<>();
        for(int i = 0; i < list1.length; i++) {
            map1.put(list1[i], i);
        }
        int min = Integer.MAX_VALUE;
        for(int i = 0; i < list2.length; i++) {
            if( map1.containsKey(list2[i]) ) {
                int index_sum = map1.get(list2[i]) + i;
                min = Math.min(index_sum, min);
                map2.put(list2[i], index_sum);
            }
        }
        List<String> list = new ArrayList<>(); 
        for(String key : map2.keySet()) {
            if(min == map2.get(key)) {
                list.add(key);
            }
        }
        return list.toArray(new String[list.size()]);
    }
}

599. Minimum Index Sum of Two Lists_字符串