Search in Rotated Sorted Array问题及解法

问题描述：

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.

问题分析：

寻找数组中一个元素是否存在，实际上就是一种查找算法。用线性查找法肯定能找到，但是时间复杂度为O(N)。因而我们采取一种二元查找法，先找出最大的元素索引，然后依次再利用二元查找法寻找target值得索引。

过程详见代码：

class Solution {
public:
    int search(vector& nums, int target) {
    	if(nums.empty()) return -1;
        int low = 0;
        int high = nums.size() - 1;
        int maxi = findMaxIndex(nums);
        if(nums[maxi] == target) return maxi;
        else if(nums[maxi] < target) return -1;
        else{
        	if(nums[low] > target) return findTarget(nums,target,maxi+1,high);
        	else return findTarget(nums,target,low,maxi);
		}
    }
    
    int findMaxIndex(vector& nums)
    {
    	int low = 0;
        int high = nums.size() - 1;
        int maxi = high; 
        if(nums[low] < nums[high]) return high;
      
		while(low < high)
	    {
	        int mid = low + (high - low) / 2;
	        if(nums[mid] > nums[mid + 1]) return mid; 
	       	else if(nums[mid] >= nums[low])
	        {
	        	low = mid + 1;
			}
			else
			{
				high = mid - 1;
			}
		}
		return low;
	}

	int findTarget(vector& nums, int target,int low, int high)
	{
        while(low < high)
        {
        	int mid = low + (high - low) / 2;
        	if(nums[mid] == target) return mid;
        	else if(nums[mid] < target) low = mid + 1;
        	else high = mid;
		}
		
		return nums[high] == target ? low : -1;
	}

};