"One Day together Leetcode" #33. Search in rotated Sorted Array

One Day together Leetcode

This series of articles has all been uploaded to my github address:
Https://github.com/Zeecoders/LeetCode
Welcome reprint, Reprint please indicate the source

(i) Title

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.

(ii) Problem solving

A well-ordered array that, after a certain amount of rotation, gets a new array, looking for a target number in this new array.
So, first we need to find the starting point of the original array in the rotated array and divide the array into two parts.
For example: 4,5,6,7,0,1,2 is divided into 4,5,6,7 and 0,1,2
Then, determine which part of the target number is,
Finally, the binary method is used to speed up the search!
See the code specifically:

Version not optimized

classSolution { Public:intSearch vector<int>& Nums,intTarget) {intLen = Nums.size ();inti =0;intj = len-1;intp =0; while(p+1<len&&nums[p]<nums[p+1]) p++;//Find the first break ascending number is the rotation center        if(nums[0]<=target) j=p;//Determine which part of the number is being searched        Else if(nums[len-1]>=target) i=p+1; while(I&LT;=J)//Two points search{intMid = (i+j)/2;if(Nums[mid] = = target)returnMidElse if(Nums[mid] >target) j= mid-1;Elsei = mid+1; }return-1;//Not found returns-1}};

After AC a look at the running time of 8ms, it seems that the code is still to be optimized, so in the search for the rotation center of the method to optimize, using the dichotomy of searching rotation center.

Optimized version

classSolution { Public:intSearch vector<int>& Nums,intTarget) {intLen = Nums.size ();inti =0;intj = Len-1;intp =0;BOOLIsfind =false;if(nums[0]>nums[len-1])//If a rotation is made{ while(I < J)//Two points search rotation center{if(i = = J-1) {Isfind =true; Break; }intMid = (i + j)/2;if(Nums[i] < Nums[mid]) i = mid;if(Nums[j] > Nums[mid]) j = mid; }        }if(Isfind)//found the rotation center{if(nums[0] <= target) {j = i; i =0; }if(Nums[len-1] >= target) {i = j; j = Len-1;} } while(i<=j) binary search target number {intMid = (i + j)/2;if(Nums[mid] = = target)returnMidElse if(Nums[mid] >target) j = mid-1;Elsei = mid+1; }return-1; }};

Optimized version run time 4ms, half faster!

"One Day together Leetcode" #33. Search in rotated Sorted Array