This problem is a nice application of binary search. The key lies in how to determine the correct half for target. Since the array has been rotated, we now need to make some additional checks.
You may run the following code with some examples to see how it works :-)
C (0ms)
1 int search(int* nums, int numsSize, int target) { 2 int l = 0, r = numsSize - 1; 3 while (l <= r) { 4 int mid = (l & r) + ((l ^ r) >> 1); 5 if (nums[mid] == target) return mid; 6 if (nums[mid] > target) { 7 if (nums[r] < target || nums[mid] < nums[r]) r = mid - 1; 8 else l = mid + 1; 9 } 10 else { 11 if (nums[r] >= target || nums[mid] >= nums[l]) l = mid + 1; 12 else r = mid - 1; 13 } 14 } 15 return -1; 16 }
C++ (4ms)
1 class Solution { 2 public: 3 int search(vector<int>& nums, int target) { 4 int l = 0, r = nums.size() - 1; 5 while (l <= r) { 6 int mid = (l & r) + ((l ^ r) >> 1); 7 if (nums[mid] == target) return mid; 8 if (nums[mid] > target) { 9 if (nums[r] < target || nums[mid] < nums[r]) r = mid - 1; 10 else l = mid + 1; 11 } 12 else { 13 if (nums[r] >= target || nums[mid] > nums[l]) l = mid + 1; 14 else r = mid - 1; 15 } 16 } 17 return -1; 18 } 19 };
Python (40ms)
1 class Solution: 2 # @param {integer[]} nums 3 # @param {integer} target 4 # @return {integer} 5 def search(self, nums, target): 6 l, r = 0, len(nums) - 1 7 while l <= r: 8 mid = (l & r) + ((l ^ r) >> 1) 9 if nums[mid] == target: 10 return mid 11 if nums[mid] > target: 12 if nums[r] < target or nums[mid] < nums[r]: 13 r = mid - 1 14 else: 15 l = mid + 1 16 else: 17 if nums[r] >= target or nums[mid] > nums[l]: 18 l = mid + 1 19 else: 20 r = mid - 1 21 return -1
[LeetCode] Search in Rotated Sorted Array
原文：http://www.cnblogs.com/jcliBlogger/p/4657810.html