[Lintcode]62. Search in Rotated Sorted Array/[Leetcode]33. Search in Rotated Sorted Array

[Lintcode]62. Search in Rotated Sorted Array/[Leetcode]33. Search in Rotated Sorted Array

• 本题难度: Medium/Medium

• Topic: Binary Search

  Description

1. Search in Rotated Sorted Array
   Suppose a sorted array 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.

Example
For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Challenge
O(logN) time

我的代码

class Solution:
    def search(self, A: 'List[int]', target: 'int') -> 'int':
        # write your code here
        if A  == []:
            return -1
    
        l = 0
        r = len(A)-1#注意！
        f = -1 # breakpoint
        while(l<=r):
            m = (l + r) // 2
            #print(l,m,r)
            if A[m]>A[0]:
                l = m+1
            else:
                #1. m == 0
                #2. m == len(A)-1 ok
    
                # 不存在需要m==0的情况
                # if m == 0:
                #     f = -1
                #     break
                if m == 0:
                    if len(A)==1:
                        f = -1
                    else:
                        if A[0]>A[1]:
                            f = 1
                        else:
                            f = -1
                    break
                if A[m-1]>A[m]:#注意向下取整
                    f =  m
                    break
                else:
                    r = m-1
        #print(f)
        if f == -1:
            l = 0
            r = len(A) - 1
        else:
            if target>=A[0]:
                l = 0
                r = f-1
            else:
                l = f
                r = len(A)-1
        while(l<=r):
            m = (l+r)//2
            if A[m] == target:
                return m
            else:
                if A[m]>target:
                    r = m-1
                else:
                    l = m+1
        return -1

别人的代码

class Solution:
    # @param {integer[]} numss
    # @param {integer} target
    # @return {integer}
    def search(self, nums, target):
        if not nums:
            return -1

        low, high = 0, len(nums) - 1

        while low <= high:
            mid = (low + high) / 2
            if target == nums[mid]:
                return mid

            if nums[low] <= nums[mid]:
                if nums[low] <= target <= nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            else:
                if nums[mid] <= target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1

        return -1

思路

情况比较多，在处理边界问题上频繁出错。
应该考虑

1. len(A)为0，1，2和大于2的情况
2. 没有翻转的情况
3. 翻转的点在1
4. 翻转的点在len(A)-1

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气死我了，看看人家的思路多好！我真是脑残。

同类问题[Leetcode]81. Search in Rotated Sorted Array II

• 时间复杂度: O(log(n))