【leetcode】Search in Rotated Sorted Array

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.

思路

二分查找，中值位置分三种情况，对这三种情况进行讨论



代码

 int search(vector<int>& nums, int target) {
        int low=0,high=nums.size()-1;
        int mid=0;
        while(low<=high)
        {
            mid=low+(high-low)/2;
            if(nums[mid]==target)return mid;//找到target
            else if(nums[mid]>nums[high])
            {
                //第二种情况
                if(target>=nums[low]&&target<nums[mid])//若target 在左边有序数组中
                    high=mid-1;
                else low=mid+1;
            }else
            {
                //第一种与第三种一起处理
                if(target>nums[mid]&&target<=nums[high])//若target在右边有序数组中
                    low=mid+1;
                else high=mid-1;
            }

        }
        return -1;
    }

Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":

What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.
第二问中包含了重复元素，做法是在上一题的基础上对于中值等于最右端值的时候最右边位置减一，继续查找。

bool search(vector<int>& nums, int target) {
//二分查找
int low=0,high=nums.size()-1;
int mid=0;
while(low<=high)
{
mid=low+(high-low)/2;
if(nums[mid]==target)return true;
else if(nums[mid]>nums[high])
{
if(target>=nums[low]&&target<nums[mid])
high=mid-1;
else low=mid+1;
}else if(nums[mid]<nums[high])
{
if(target>nums[mid]&&target<=nums[high])
low=mid+1;
else high=mid-1;
}else
{

high--;//这个地方降低了查找效率
}

}
return false;
}