Leetcode 33-search in rotated Sorted Array (c + + Java Python)

Title: http://oj.leetcode.com/problems/search-in-rotated-sorted-array/

Suppose a sorted array is rotated in some pivot unknown to your beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are 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.

Title translation:

Suppose an ordered array rotates in an unknown position.
(i.e. 0 1 2 4 5 6 7 May become 4 5 6 7 0 1 of 2).
Searches for a given target value. Returns the index of the array if it exists, or returns-1.
You can assume that there are no duplicate elements in the array.
Analysis:
Similar to a binary lookup, but it is more complicated to determine which child interval to be in.
C + + implementation:

Class Solution {public
:
    int search (int a[], int n, int target) {
    	int left = 0;
    	int right = n-1;

    	While [left <= right]
    	{
    		int mid = (left + right)/2;

    		if (a[mid] = = target)
    		{return
    			mid;
    		}

    		if (A[mid] >= A[left])
    		{
    			if (A[mid] > Target && a[left] <= target)
				{right
					= mid-1;
  }
				Else
				{left
					= mid + 1;
				}
    		}
    		else
    		{
    			if (A[mid] < target && A[right] >= target)
    			{left
    				= mid + 1;
    			}
    			else
    			{Right
    				= mid-1

    	}}} return-1;
    }
;

Java implementation:

public class Solution {public
    int search (int[] A, int target) {
		int left = 0;
		int right = A.length-1;

		While [left <= right] {
			int mid = (left + right)/2;

			if (a[mid] = = target) {return
				mid;
			}

			if (A[mid] >= A[left]) { 
				if (A[mid] > Target && a[left] <= target) {right
					= mid-1;
				} els e {Left
					= mid + 1;
				}
			} else {
				if (A[mid] < target && A[right] >= target) {left
					= Mid + 1;
				} else {Right
					= mid-1

		}}} Return-1
    }
}

Python implementations:

Class Solution:
    # @param A, a list of integers
    # @param target, an integer to is searched
    # @return an integer
    def search (self, A, target): Left
        = 0 Right
        = Len (A)-1 while left
        
        <= right:
            mid = (left + right) /2
            
            if a[mid] = = target: Return
                mid
            
            if A[mid] >= A[left]:
                if A[MID] > Target and a[left] <= ta Rget: Right
                    = mid-1
                else: Left
                    = mid + 1
            else:
                if A[MID] < target and A[right] >= target: Left
                    = mid + 1
                else: Right
                    = mid-1
            
        return-1

Thank you for reading and welcome comments.