原题链接在这里:https://leetcode.com/problems/minimum-index-sum-of-two-lists/description/

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:

  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.

题解:

把list1的element 和对应的index放进HashMap<String, Integer> hm中. 再iterate list2, 如果list2的element在hm中比较index相加是否比minIndexSum小,若比minIndexSum小,清空res重新加, 更新minIndexSum. 若相等直接加进res中.

Time Complexity: O(list1.length + list2.length)

Space: O(list1.length).

AC Java:

 class Solution {
public String[] findRestaurant(String[] list1, String[] list2) {
List<String> res = new ArrayList<String>();
int minIndexSum = Integer.MAX_VALUE;
HashMap<String, Integer> hm = new HashMap<String, Integer>();
for(int i = 0; i<list1.length; i++){
hm.put(list1[i], i);
} for(int j = 0; j<list2.length; j++){
if(hm.containsKey(list2[j])){
int i = hm.get(list2[j]);
if(i+j < minIndexSum){
res.clear();
res.add(list2[j]);
minIndexSum = i+j;
}else if(i+j == minIndexSum){
res.add(list2[j]);
}
}
} return res.toArray(new String[res.size()]);
}
}

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