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.

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寻找最小值时关键在于 寻找，两个递增数组，第一个最大右边界，第二个最小左边界

class Solution {public:    int bs(vector<int>& nums, int st, int en, int target)    {    if(st > en)    return -1;    int m;    while(st <= en)    {    m = st + (en - st)/2;    if(nums[m] == target)    {    return m;    }    if(nums[m] < target)    {    st = m + 1;    }else    {    en = m - 1;    }    }    return -1;    }    int search(vector<int>& nums, int target)    {    int l = 0;    int r = nums.size() - 1;    int m;    int pos = -1;    if(nums[l] < nums[r])    {    pos = 0;    return bs(nums, 0, r, target);    }    while(l < r)        {    if(r - l == 1)    {    pos = l;    break;    }    m = l + (r - l)/2;    if(nums[m] > nums[l])    {    l = m;    }else if(nums < nums[r])    {    r = m;    }    }    int lpos = bs(nums, 0, pos - 1, target);    int rpos = bs(nums, pos, nums.size() - 1, target);    if(lpos >= 0) return lpos;    if(rpos >= 0) return rpos;    return -1;    }};