33. Search in Rotated Sorted Array

题目

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.
分析
在旋转数组中查找给定元素，可以先找到旋转点，然后再用二分法分别从前半部分和后半部分去查找，或者直接进行二分查找，对mid值和target还有beg，end等情况进行分析。

先找旋转点再二分：

class Solution {
public:
int findDiv(int beg,int end,vector<int>& nums){//寻找旋转点下标，如果是未旋转返回-1
int mid=-1;
while(beg<end){//因为mid取beg+end中间值的天棚，即如果beg和end相邻，则mid取为end，所以循环中止条件为beg<end
mid=(beg+end+1)/2;//因为当nums[mid]<=nums[beg]时，end=mid，所以如果不取天棚，则当beg与end相邻，且end为旋转点，则mid=beg，下一步end=mid=beg，退出循环，无法返回旋转点；倘若循环后加入对beg=end的判断，则无法区分此情况与未旋转的情况，因为未旋转时，beg会一直增到end前
if(nums[mid]>nums[beg])
beg=mid;
else{
if(nums[mid]<nums[mid-1])
return mid;
else
end=mid;
}
}
return -1;
}
int binSearch(int beg,int end,vector<int>& nums,int target){//二分查找target
while(beg<=end){
int mid=(beg+end)/2;
if(nums[mid]==target)
return mid;
else if(nums[mid]<target)
beg=mid+1;
else
end=mid-1;
}
return -1;
}
int search(vector<int>& nums, int target) {
if(nums.size()<3){//对于规模较小的，直接比较
for(int i=0;i<nums.size();++i){
if(nums[i]==target)
return i;
}
}
int div=findDiv(0,nums.size()-1,nums);//找到旋转点
if(div==-1){//代表序列未旋转，正常二分
return binSearch(0,nums.size()-1,nums,target);
}
else{//否则根据div值分为左右两部分进行二分，其中div是后半序列的首元素下标
int left=binSearch(0,div-1,nums,target);
int right=binSearch(div,nums.size()-1,nums,target);
return left==-1?right:left;
}
return -1;
}
};直接二分：
class Solution {
public:
int search(vector<int>& nums, int target) {
if(nums.size() == 0)
return -1;
if(nums.size() == 1)
return target == nums[0]? 0: -1;
int begin = 0;
int end = nums.size() - 1;
while(begin < end)
{
int mid_ind = (begin + end)/2;
if (target == nums[mid_ind])
return mid_ind;
if (target == nums[end])
return end;
if (target == nums[begin])
return begin;

if(nums[begin] < nums[mid_ind]) //ascending
{

if(target < nums[mid_ind] && target >= nums[begin])
end = mid_ind -1;
else
begin = mid_ind+1;

}
else //if(nums[mid_ind] < nums[end])
{
if(target > nums[mid_ind] && target <= nums[end])
begin = mid_ind +1;
else
end = mid_ind -1;
}

}
return -1;
}
};