LeetCode 33. Search in Rotated Sorted Array

题意：

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.

思路：

先找到分界点。找分界点参考：http://www.cnblogs.com/gufeiyang/p/5291679.html

 

找分界点，二分的时候边界一定要处理好。

找到边界后，选择一部分继续二分找到答案。

AC代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        if(n == 0) return -1;
        int L = 0, R = n-1;
        if(nums[0] < nums[n-1])
        {
            L = 0;
            R = n-1;
        }
        else 
        {
            int left = 0, right = n-1;
            while(left<right)
            {
                int mid = (left+right)/2;
                if(nums[mid]>=nums[0]) left=mid+1;
                else if(nums[mid] < nums[0]) right = mid;
            }
            while(left>0 && nums[left]>nums[left-1]) left--;
            if(nums[n-1] >= target)
            {
                L = right;
                R = n-1;
            }
            else 
            {
                L = 0;
                R = right-1;
            }
        }
        while(L <= R)
        {
            int mid = (L+R)/2;
            if(nums[mid] == target) return mid;
            else if(nums[mid] < target) L = mid+1;
            else R = mid-1;
        }
        return -1;
    }
};