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：

 1 class Solution {
 2 public:
 3     int search(vector<int>& nums, int target) {
 4         int pivot = getPivot(nums, 0, nums.size() - 1);
 5         int pos = bSearch(nums, 0, pivot - 1, target);
 6         if(pos != -1)
 7             return pos;
 8         return bSearch(nums, pivot, nums.size() - 1, target);
 9     }
10     
11     int getPivot(vector<int>&nums, int start, int end){
12         if(start > end)
13             return -1;
14         int mid = start + (end - start)/2;
15         if(nums[start] <= nums[mid]){
16             int pos = getPivot(nums, mid + 1, end);
17             if(pos == -1) return start;
18             else
19                 if(nums[pos] < nums[start])
20                     return pos;
21                 else
22                     return start;
23         }else{
24             int pos = getPivot(nums, start, mid);
25             if(pos == -1)
26                 return mid;
27             if(nums[pos] < nums[end])
28                 return pos;
29             else
30                 return mid;
31         }
32     }
33     
34     int bSearch(vector<int>&nums, int start, int end, int target){
35        int mid;
36        while(start <= end){
37            mid = (start+end)/2;
38            if(nums[mid] > target){
39                end = mid - 1;
40            }else if(nums[mid] < target){
41                start = mid + 1;
42            }else{
43                return mid;
44            }
45        } 
46        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
47     }
48 };