https://leetcode.com/problems/search-in-rotated-sorted-array/
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https://leetcode.com/problems/search-in-rotated-sorted-array/
1.寻找初始的index 也就是rotate的情况
2. 在顺序的列表里 二分查找
如果没有 返回-1 如果有 再减去startindex
注意 减去startindex可能会出现负数的情况 在加一个len(list)再取len(list)余
class Solution: # @param {integer[]} nums # @param {integer} target # @return {integer} def search(self, nums, target): #找startindex并且把nums编程普通顺序的 startindex=0 for i in range(len(nums)-1): if nums[i]>nums[i+1]: startindex=len(nums)-1-i nums=nums[i+1:]+nums[:i+1] s=0 e=len(nums)-1 #过大过小 if target<nums[0] or target>nums[-1]: return -1 #和首尾相同 if target==nums[-1] : return len(nums)-1-startindex if target==nums[0] : return (len(nums)-startindex)%len(nums) 二分查找 while s!=e-1: if nums[(e+s)/2]>target: e=(e+s)/2 elif nums[(e+s)/2]<target: s=(e+s)/2 else: return ((e+s)/2-startindex+len(nums))%len(nums) return -1
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