[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).

思路：借用一个Map和Set, map的key为String, value为String在list1和list2中出现的索引。 Set中盛放最终结果，再把Set转化输出需要的String[]

package com.billkang;import java.util.HashMap;import java.util.HashSet;import java.util.Map;import java.util.Set;/** * @author binkang * @date May 28, 2017 */public class MinimumIndexSumofTwoLists {public String[] findRestaurant(String[] list1, String[] list2) {Map<String, Integer> map = new HashMap<String, Integer>();for(int i=0;i<list1.length;i++) {map.put(list1[i], i);}Set<String> set = new HashSet<String>();int min = Integer.MAX_VALUE;for(int j=0;j<list2.length;j++) {if(map.get(list2[j])!=null) {int i = map.get(list2[j]);map.put(list2[j], i+j);if(i+j==min) {set.add(list2[j]);} else if(i+j<min) {set.clear();set.add(list2[j]);min = i+j;}}}String[] result = new String[set.size()];int i=0;for(String s : set) {result[i++] = s;}return result;}}