Search in Rotated Sorted Array II

描述

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.

中文：

如果允许重复元素，怎么办

分析：

重复元素的话，主要是

当nums[first]==nums[mid]时，无法判断元素的序列是否递增，也无法找到转折点如 131111111怎么找？

所以要单独分析，那就只能将first向前推进。

class Solution {
public:
bool search(vector<int>& nums, int target) {
int first=0;
int last=nums.size()-1;
int mid;

while(first<=last)
{
mid=(first+last)/2;
if(nums[mid]==target)
return true;
if(nums[first]<nums[mid])
{
if(target>=nums[first]&&target<nums[mid])
last=mid-1;
else
first=mid+1;
}
else if(nums[first]>nums[mid])
{
if(target>nums[mid]&&target<=nums[last])
first=mid+1;
else
last=mid-1;
}
else
first++;
}

return false;
}
};

Runtime: 8 ms