LeetCode OJ：Search in Rotated Sorted Array（翻转排序数组的查找）

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.

相当于将数组的后面一部折叠到数组的前面去了，本质上也是二分法，这里其实也是有规律可循的：首先要得到转折点，也就是其左边右边都比其大的那一点。给一个start以及一个end，如果nums[mid] > nums[start]，那么说明从start到mid也一定是递增的，很容的知道pivot一定在mid+1到end之间，那么递归的去查找就可以了。nums[mid] < nums[start]可以同样的去分析，代码如下所示：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int pivot = getPivot(nums, 0, nums.size() - 1);
        int pos = bSearch(nums, 0, pivot - 1, target);
        if(pos != -1)
            return pos;
        return bSearch(nums, pivot, nums.size() - 1, target);
    }
    
    int getPivot(vector<int>&nums, int start, int end){
        if(start > end)
            return -1;
        int mid = start + (end - start)/2;
        if(nums[start] <= nums[mid]){
            int pos = getPivot(nums, mid + 1, end);
            if(pos == -1) return start;
            else
                if(nums[pos] < nums[start])
                    return pos;
                else
                    return start;
        }else{
            int pos = getPivot(nums, start, mid);
            if(pos == -1)
                return mid;
            if(nums[pos] < nums[end])
                return pos;
            else
                return mid;
        }
    }
    
    int bSearch(vector<int>&nums, int start, int end, int target){
       int mid;
       while(start <= end){
           mid = (start+end)/2;
           if(nums[mid] > target){
               end = mid - 1;
           }else if(nums[mid] < target){
               start = mid + 1;
           }else{
               return mid;
           }
       } 
       return -1; //返回-1，表明在这一部分没有找到，可以在下一部分查找
    }
};