C#LeetCode刷题之#33-搜索旋转排序数组(Search in Rotated Sorted Array)

C#LeetCode刷题之#33-搜索旋转排序数组(Search in Rotated Sorted Array)

问题

假设按照升序排序的数组在预先未知的某个点上进行了旋转。

( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。

搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 。

你可以假设数组中不存在重复的元素。

你的算法时间复杂度必须是 O(log n) 级别。

输入: nums = [4,5,6,7,0,1,2], target = 0

输出: 4

输入: nums = [4,5,6,7,0,1,2], target = 3

输出: -1

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

Input: nums = [4,5,6,7,0,1,2], target = 0

Output: 4

Input: nums = [4,5,6,7,0,1,2], target = 3

Output: -1

示例

public class Program {

    public static void Main(string[] args) {
        int[] nums = { 2, 7, 11, 15 };

        var res = TwoSum(nums, 9);
        var res2 = TwoSum2(nums, 9);

        Console.WriteLine($"[{res[0]},{res[1]}]");
        Console.WriteLine($"[{res2[0]},{res2[1]}]");

        Console.ReadKey();
    }

    public static int[] TwoSum(int[] nums, int target) {
        //类似于冒泡,双循环比较2个数的和与目标值是否相等
        for (int i = 0; i < nums.Length; i++) {
            for (int j = i + 1; j < nums.Length; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[] { i, j };
                }
            }
        }
        //找不到时抛出异常
        throw new ArgumentException();
    }

    public static int[] TwoSum2(int[] nums, int target) {
        //用数组中的值做key,索引做value存下所有值
        var dictionary = new Dictionary<int, int>();
        for (int i = 0; i < nums.Length; i++) {
            //记录差值
            int complement = target - nums[i];
            //若字典中已经存在这个值,说明匹配成功
            if (dictionary.ContainsKey(complement)) {
                return new int[] { dictionary[complement], i };
            }
            //记录索引
            dictionary[nums[i]] = i;
        }
        //找不到时抛出异常
        throw new ArgumentException();
    }

}

以上给出2种算法实现,以下是这个案例的输出结果:

[0,1]
[0,1]

分析

显而易见,TwoSum 的时间复杂度为: O(n^{2})​ ,TwoSum2 的时间复杂度为: O(n)​ 。

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