C#LeetCode刷题之#599-两个列表的最小索引总和​​​​​​​​​​​​​​(Minimum Index Sum of Two Lists)

C#LeetCode刷题之#599-两个列表的最小索引总和​​​​​​​​​​​​​​(Minimum Index Sum of Two Lists)

问题

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。

输入:
[“Shogun”, “Tapioca Express”, “Burger King”, “KFC”]
[“Piatti”, “The Grill at Torrey Pines”, “Hungry Hunter Steakhouse”, “Shogun”]

输出: [“Shogun”]

解释: 他们唯一共同喜爱的餐厅是“Shogun”。

输入:
[“Shogun”, “Tapioca Express”, “Burger King”, “KFC”]
[“KFC”, “Shogun”, “Burger King”]

输出: [“Shogun”]

解释: 他们共同喜爱且具有最小索引和的餐厅是“Shogun”,它有最小的索引和1(0+1)。

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。

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.

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”.

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:

The length of both lists will be in the range of [1, 1000].
The length of strings in both lists will be in the range of [1, 30].
The index is starting from 0 to the list length minus 1.
No duplicates in both lists.

示例

public class Program {

    public static void Main(string[] args) {
        var list1 = new string[] { "Shogun", "Tapioca Express", "Burger King", "KFC" };
        var list2 = new string[] { "KFC", "Shogun", "Burger King" };

        var res = FindRestaurant(list1, list2);
        ShowArray(res);

        Console.ReadKey();
    }

    private static void ShowArray(string[] array) {
        foreach(var interest in array) {
            Console.Write($"{interest} ");
        }
        Console.WriteLine();
    }

    public static string[] FindRestaurant(string[] list1, string[] list2) {
        var min = int.MaxValue;
        var dic = new Dictionary<string, int>();
        for(var i = 0; i < list1.Length; i++) {
            for(var j = 0; j < list2.Length; j++) {
                if(list1[i] == list2[j]) {
                    min = Math.Min(min, i + j);
                    dic.Add(list1[i], i + j);
                }
            }
        }
        return (from r in dic.ToList()
                where r.Value == min
                select r.Key).ToArray();
    }

}

以上给出1种算法实现,以下是这个案例的输出结果:

Shogun

分析:

显而易见,以上算法的时间复杂度为: O(m*n)

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