33 Search in Rotated Sorted Array 搜索旋转排序数组
Description:
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).
Example:
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
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 :
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
思路:
- 暴力法, 逐个查找
时间复杂度O(n), 空间复杂度O(1) - 旋转之后的数组也是部分有序的, 可以用类似二分查找的方法查找
时间复杂度O(lgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int search(vector& nums, int target)
{
int low = 0, high = nums.size() - 1;
while (low <= high)
{
int mid = ((high - low) >> 1) + low;
if (nums[mid] == target) return mid;
else if (nums[mid] < nums[high])
{
if (nums[mid] < target and target <= nums[high]) low = mid + 1;
else high = mid - 1;
}
else
{
if (nums[mid] > target and target >= nums[low]) high = mid - 1;
else low = mid + 1;
}
}
return -1;
}
};
Java:
class Solution {
public int search(int[] nums, int target) {
int result = -1;
for (int i = 0; i < nums.length; i++) if (nums[i] == target) result = i;
return result;
}
}
Python:
class Solution:
def search(self, nums: List[int], target: int) -> int:
return nums.index(target) if target in nums else -1