[LeetCode] 599. Minimum Index Sum of Two Lists

轉自LeetCode

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.

<Solution>

想法如下

• 先用一個 HashMap 將 list 1 的值都存起來，key 是 String 然後 value 是 index

• 歷遍 list2，檢查該 String 在不在 map 裡面，如果有，計算目前的 index sum。若是 index sum 比目前的index sum的更小，清空之前的記錄的答案，更新最小值，重新紀錄答案 ; 如果和目前的index sum 一樣大，那就把該 string 放到答案中

code 如下

Java

	class Solution {
	public String[] findRestaurant(String[] list1, String[] list2) {
	HashMap<String, Integer> map = new HashMap<>();
	for(int i = 0; i < list1.length; i++) {
	map.put(list1[i], i);
	}
	ArrayList<String> ans = new ArrayList<>();
	int minSum = Integer.MAX_VALUE;
	int tmpSum;
	for(int i = 0; i < list2.length && i <= minSum; i++) {
	if(map.containsKey(list2[i])) {
	tmpSum = i + map.get(list2[i]);
	if(tmpSum < minSum) {
	minSum = tmpSum;
	ans.clear();
	ans.add(list2[i]);
	}
	else if(tmpSum == minSum) {
	ans.add(list2[i]);
	}
	}
	}
	return ans.toArray(new String[0]);
	}
	}

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