leetcode 33:Revised Binary Search

题目：

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.

class Solution {public:    int search(vector<int>& nums, int target) {            }};

思路：

虽然不是所有元素都排序了的，但至少可以分成两部分，且两部分都是排好序的。因此我们还是可以用binary search方法来解。

但是怎样来进行取舍呢，即不能简单的根据target与中间元素的大小来决定是搜索左半边还是右半边。

注意到，不管上面的怎样旋转，我们取定中间元素后，一定有半边是严格排序的，比如：

6 7 0 1 2 4 5，我们取中间元素1，则右边是严格排序的，同样地

2 4 5 6 7 0 1，我们取中间元素6，则左边是严格排序的。

所以只需判断target是否在严格排序那边即可。

代码看起来比较复杂，其实很好理解，时间复杂度：O(lgn)

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