0599. Minimum Index Sum of Two Lists

599. Minimum Index Sum of Two Lists #

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.

题目大意 #

假设 Andy 和 Doris 想在晚餐时选择一家餐厅,并且他们都有一个表示最喜爱餐厅的列表,每个餐厅的名字用字符串表示。你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个,则输出所有答案并且不考虑顺序。 你可以假设总是存在一个答案。

提示:

  • 两个列表的长度范围都在 [1, 1000] 内。
  • 两个列表中的字符串的长度将在 [1,30] 的范围内。
  • 下标从 0 开始,到列表的长度减 1。
  • 两个列表都没有重复的元素。

解题思路 #

  • 在 Andy 和 Doris 两人分别有各自的餐厅喜欢列表,要求找出两人公共喜欢的一家餐厅,如果共同喜欢的次数相同,都输出。这一题是简单题,用 map 统计频次,输出频次最多的餐厅。

代码 #


package leetcode

func findRestaurant(list1 []string, list2 []string) []string {
	m, ans := make(map[string]int, len(list1)), []string{}
	for i, r := range list1 {
		m[r] = i
	}
	for j, r := range list2 {
		if _, ok := m[r]; ok {
			m[r] += j
			if len(ans) == 0 || m[r] == m[ans[0]] {
				ans = append(ans, r)
			} else if m[r] < m[ans[0]] {
				ans = []string{r}
			}
		}
	}
	return ans
}


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