33. Search in Rotated Sorted Array

/*Suppose an array sorted in ascending order 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.思路：    1.使用二分法找到最大值与最小值跳变沿 ，记录最小值的索引为start2.在索引[0,start)半开半闭区间内为增序列 在[start,numsSize-1]区间内为增序列3.再次使用二分查找找到target值*/#include <stdio.h>int search(int* nums, int numsSize, int target){    if(numsSize==0)return -1;int mid=0,left=0,right=numsSize-1;while(left<right){mid=(left+right)>>1;if(nums[mid]>=nums[right])left=mid+1;elseright=mid;}int start=left;left=0,right=numsSize-1;int realmid=0;while(left<=right){mid=(left+right)>>1;realmid=(mid+start)%numsSize;//以start为升序数组的起点，寻找真正的中心点if(nums[realmid]==target) return realmid;if(nums[realmid]<target) left=mid+1;else right=mid-1;}return -1;}int search2(int* nums, int numsSize, int target){    if(numsSize==0)return -1;int mid=0,left=0,right=numsSize-1;while(left<right){mid=(left+right)>>1;if(nums[mid]>=nums[right])left=mid+1;elseright=mid;}int start=left;left= target<=nums[numsSize-1] ? start : 0;right= target>nums[numsSize-1] ? start-1 : numsSize-1;while(left<=right){mid=(left+right)>>1;if(nums[mid]==target) return mid;if(nums[mid]<target) left=mid+1;else right=mid-1;}return -1;}int main(){int arr[]={4,5,6,7,0,1,2};int len=sizeof(arr)/sizeof(int);printf("%d\n",search2(arr,len,5));return 0;}