Leetcode: Search in rotated sorted Array)

Question:

Suppose a sorted array is rotated at some unknown to you beforehand.

(I. e .,0 1 2 4 5 6 7Might become4 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.

Note:

1) binary search for sorted Arrays

2) locate critical points

Implementation:

I. My code:

1 class solution {2 public: 3 int search (int A [], int N, int target) {4 If (n = 0 | n = 1 & A [0]! = Target) Return-1; 5 if (a [0] = target) return 0; 6 int I = 1; 7 while (A [I-1] <A [I]) I ++; 8 9 int pre = binary_search (A, 0, I, target); 10 int Pos = binary_search (A, I, n-I, target ); 11 return pre =-1? POs: Pre; 12} 13 private: 14 int binary_search (int * B, int Lo, int Len, int goal) 15 {16 int low = lo; 17 int high = lo + len-1; 18 while (low <= high) 19 {20 int middle = (low + high)/2; 21 if (Goal = B [Middle]) // find and return index22 return middle; 23 else if (B [Middle] <goal) // on the right side 24 low = middle + 1; 25 else // on the left side 26 high = middle-1; 27} 28 return-1; // No, -129} 30 };

2. Open source code on the Internet:

 

 1 class Solution { 2 public: 3     int search(int A[], int n, int target) { 4         int first = 0, last = n-1; 5         while (first <= last)  6         { 7             const int mid = (first + last) / 2; 8             if (A[mid] == target) 9                 return mid;10             if (A[first] <= A[mid]) 11             {12                 if (A[first] <= target && target < A[mid])13                    last = mid-1;14                 else15                    first = mid + 1;16             } 17             else 18             {19                 if (A[mid] < target && target <= A[last])20                     first = mid + 1;21                 else22                     last = mid-1;23             }24         }25         return -1;26     }27 };