[Array]033 Search in Rotated Sorted Array

• 分类：Array

• 考察知识点：Array(数组遍历) 二分法

• 最优解时间复杂度：O(logn)

33. Search in Rotated Sorted Array

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).

Example1:

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

Example2:

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

代码：

我的Python取巧解法:

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        #就4行，简单到你想笑晕在厕所
        try:
            return nums.index(target)
        except:
            return -1

正常解法

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        #判定边界条件
        if len(nums)==0:
            return -1
        
        start=0
        end=len(nums)-1
        while(end>start+1):
           #判定是end>start+1!!!
            mid=(end-start)//2+start
            if nums[mid]==target:
                return mid
            if(nums[start]

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讨论：

1.看到O(logn)的时间复杂度，没有什么可说，就是用两个指针然后2分法
2.用Python自带的方法时间复杂度是O(n),所以慢，但是写起来真的超级简单
3.start+1是这个判定条件，不是别的，千万别写错了

python自带方法

传统解法