LeetCode 33 Search in Rotated Sorted Array (C,C++,Java,Python)

Problem:

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.

Solution:

本题是二分查找的变形题，主要就是找到翻转的那个临界点（下一个数值比它小）就可以了，然后看target的大小判断应该在哪个范围内进行二分查找。

题目大意：

给一个排好序的数组，但是被翻转过了，就是前边的一部分被接到了数组后边，现在给一个目标值，要求得到该值在数组中的下标，没有输出-1

Java源代码（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语言源代码（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++源代码（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源代码（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