leetcode 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.

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.

我的思路就是两个指针，一个从左到右，还有一个来回指，找相同的字符串。

package leetcode;import java.util.ArrayList;import java.util.Arrays;public class Minimum_Index_Sum_of_Two_Lists_599 {public String[] findRestaurant(String[] list1, String[] list2) {ArrayList<String> resultlist=new ArrayList<String>();int resultCount=0;int minSum=Integer.MAX_VALUE;int pointer1=0;int pointer2=0;for(pointer1=0;pointer1<list1.length;pointer1++){pointer2=0;if((pointer1+pointer2)>minSum){break;}while (pointer2 < list2.length && (pointer1 + pointer2) <= minSum) {if (list1[pointer1].equals(list2[pointer2])) {if((pointer1 + pointer2)<minSum){minSum=pointer1 + pointer2;resultCount=1;resultlist.clear();resultlist.add(list1[pointer1]);}else{//(pointer1 + pointer2)=minSumresultCount++;resultlist.add(list1[pointer1]);}break;} else {pointer2++;}}}String[] result=new String[resultCount];for(int i=0;i<resultCount;i++){result[i]=resultlist.get(i);}return result;}public static void main(String[] args) {// TODO Auto-generated method stubMinimum_Index_Sum_of_Two_Lists_599 m=new Minimum_Index_Sum_of_Two_Lists_599();String[] s1=new String[]{"Shogun", "Tapioca Express", "Burger King", "KFC"};String[] s2=new String[]{"Piatti","The Grill at Torrey Pines","Hungry Hunter Steakhouse","Shogun"};String[] s3=new String[]{"KFC","Burger King","Tapioca Express","Shogun"};System.out.println(Arrays.toString(m.findRestaurant(s1, s2)));}}

大神则使用map，使得解法 O(n+m) Time O(n) Space。

public String[] findRestaurant(String[] list1, String[] list2) {    Map<String, Integer> map = new HashMap<>();    List<String> res = new LinkedList<>();    int minSum = Integer.MAX_VALUE;    for (int i=0;i<list1.length;i++){    map.put(list1[i], i);    }    for (int i=0;i<list2.length;i++) {        Integer j = map.get(list2[i]);        if (j != null && i + j <= minSum) {            if (i + j < minSum) {             res = new LinkedList<>();             minSum = i+j;             }            res.add(list2[i]);        }    }    return res.toArray(new String[res.size()]);}