Leetcode Search in rotated Sorted Array (C,c++,java,python)

Problem:

Suppose a sorted array is rotated on 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 is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

Solution: The problem is a two-point search of the deformation, mainly to find the tipping point (the next value is smaller than it) on it, and then look at the target size to determine in which range to find the binary.
The main idea: to a well-ordered array, but was flipped over, is a part of the front is received by the array, now give a target value, ask for the value in the array subscript, no output-1
Java source Code (329MS):

public class Solution {public    int search (int[] nums, int target) {        int index=0,len=nums.length;        while (index<len-1 && nums[index]<=nums[index+1]) index++;        if (Target>=nums[0] && Target<=nums[index]) {            return find (Nums,0,index,target);        } else{            return Find (Nums,index+1,len-1,target);        }    }    private int Find (int[] nums,int start,int end,int target) {        if (start>end) return-1;        int l=start,r=end,mid;        while (l<=r) {            mid= (l+r)/2;            if (Nums[mid]==target) return mid;            else if (Target<nums[mid]) r=mid-1;            else l=mid+1;        }        return-1;}    }

C Language Source code (3MS):

int find (int* nums,int start,int end,int target) {    int l=start,r=end,mid;    if (start>end) return-1;    while (l<=r) {        mid= (l+r) >>1;        if (Nums[mid]==target) return mid;        else if (nums[mid]>target) r=mid-1;        else l=mid+1;    }    return-1;} int search (int* nums, int numssize, int target) {    int index=0;    while (Index<numssize-1 && nums[index]<=nums[index+1]) index++;    if (target >= nums[0] && Target<=nums[index]) {        return find (Nums,0,index,target);    } else{        return Find (Nums,index+1,numssize-1,target);}    }

C + + source code (6MS):

Class Solution {public:    int search (vector<int>& nums, int target) {        int index=0,len=nums.size ();        while (index<len-1 && nums[index]<=nums[index+1]) index++;        if (Target>=nums[0] && Target<=nums[index]) {            return find (Nums,0,index,target);        } else{            return Find (Nums,index+1,len-1,target);        }    } Private:    int Find (vector<int>& nums,int start,int end,int target) {        if (start>end) return-1;        int l=start,r=end,mid;        while (l<=r) {            mid= (l+r) >>1;            if (Nums[mid]==target) return mid;            else if (nums[mid]>target) r=mid-1;            else l=mid+1;        }        return-1;    }};

Python source code (64MS):

Class solution:    # @param {integer[]} nums    # @param {integer} target    # @return {integer}    def search (self, Nums, target):        Index=0;length=len (nums) while        index<length-1 and Nums[index]<nums[index+1]:index+=1        if target>=nums[0] and Target<=nums[index]:return self.find (nums,0,index,target)        Else:return Self.find (nums,index+1,length-1,target)    def find (self,nums,start,end,target):        if start>end:return-1;        L=start;r=end while        l<=r:            mid= (l+r)/2            if Nums[mid]==target:return mid            elif target>nums[ Mid]:l=mid+1            else:r=mid-1        return-1

Leetcode Search in rotated Sorted Array (C,c++,java,python)