0033.Search in Rotated Sorted Array

33. 搜索旋转排序数组

English Version

题目描述

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。

示例 1:

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

示例 2:

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

解法

二分查找。

Python3

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1
        while l <= r:
            mid = (l + r) >> 1
            if nums[mid] == target:
                return mid
            if nums[mid] > target:
                if nums[mid] >= nums[r] and target < nums[l]:
                    l = mid + 1
                else:
                    r = mid - 1
            else:
                if nums[mid] <= nums[l] and target > nums[r]:
                    r = mid - 1
                else:
                    l = mid + 1
        return -1

Java

class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = (l + r) >>> 1;
            if (nums[mid] == target) return mid;
            if (nums[mid] > target) {
                if (nums[mid] >= nums[r] && target < nums[l]) l = mid + 1;
                else r = mid - 1;
            } else {
                if (nums[mid] <= nums[l] && target > nums[r]) r = mid - 1;
                else l = mid + 1;
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size() - 1;
        while (l <= r) {
            int mid = (l + r) >> 1;
            if (nums[mid] == target) return mid;
            if (nums[mid] > target) {
                if (nums[mid] >= nums[r] && target < nums[l]) l = mid + 1;
                else r = mid - 1;
            } else {
                if (nums[mid] <= nums[l] && target > nums[r]) r = mid - 1;
                else l = mid + 1;
            }
        }
        return -1;
    }
};

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