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.

分析

二分查找，

首先判断中间数是否为所求，是则返回，否则继续；

判断中间数是属于哪个序列，是左递增序列，还是右递增序列：

左递增：

    判断target是否在左递增序列中，是则更新 R = M - 1;

    否则 L = M + 1；

右递增：

    判断target是否在右递增序列中，是则更新 L = M + 1;

    否则 R = M - 1;

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代码

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class Solution { public:     int search(vector<int>& nums, int target) {         int l = 0, r = nums.size() - 1, m, result = -1;         while(l <= r){             m = (l + r) >> 1;             if(nums[m] == target)                 return m;             if(nums[l] <= nums[m])// indicate that the nums[m] is in the ascending order part             {                 if( nums[l] <= target && nums[m] > target){// target is in [l,m)                     r = m - 1;                 }                 else{ //target is in [m,R)                     l = m + 1;                 }             }             else{                 if(nums[m] < target && nums[r] >= target){ //target is in (m,r]                     l = m + 1;                 }                 else{// target is in [l,m)                     r = m - 1;                 }             }         }//end of while         return -1;     } };

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