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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 };
LeetCode OJ：Search in Rotated Sorted Array（翻转排序数组的查找）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/5106656.html
