leetcode 81. Search in Rotated Sorted Array II 旋转数组查找+二分查找

Follow up for “Search in Rotated Sorted Array”:

What if duplicates are allowed?

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

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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

这道题很简单，直接遍历即可。也可以使用二分查找。

代码如下：

public class Solution
{
/*
* 最简单的就是遍历
* */
public boolean search(int[] nums, int target)
{
if(nums==null || nums.length<=0)
return false;
for(int i=0;i<nums.length;i++)
if(nums[i]==target)
return true;
return false;
}

/*
* 二分查找
* */
public boolean search2(int[] nums, int target)
{
if(nums==null || nums.length<=0)
return false;

int left=0,right=nums.length-1;
while(left<=right)
{
int mid = (left-right)/2+right;
if(nums[mid]==target)
return true;

if(nums[left]<nums[mid])
{
if(nums[left]<=target&& target<nums[mid])
right=mid-1;
else
left=mid+1;
}else if(nums[left]>nums[mid])
{
if(nums[mid]<target&&target<=nums[right])
left=mid+1;
else
right=mid-1;
}else
left++;
}

return false;
}

}

下面是C++的做法，我是直接遍历

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>

using namespace std;

class Solution
{
public:
bool search(vector<int>& a, int target)
{
if (a.size() <= 0)
return false;
int left = 0, right = a.size() - 1;
while (left <= right)
{
int mid = (right - left) / 2 + left;
if (a[mid] == target)
return true;
else  if (a[left] < a[mid])
{
if (a[left] <= target && target < a[mid])
right = mid - 1;
else
left = mid + 1;
}
else if (a[left] > a[mid])
{
if (a[mid] < target && target <= a[right])
left = mid + 1;
else
right = mid - 1;
}
else
left++;
}
return false;
}
};