[LeetCode]26. Search in Rotated Array II旋转数组查找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.

解法：同旋转数组查找I，本题也使用二分查找，只是在nums[mid]和nums[right]（或者nums[right]）比较时若二者相等则left++（或者right--）即可。nums[mid]与nums[left]比较的代码：

class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
bool res = false;
int left = 0, right = n - 1;

while(left <= right)
{
int mid = ((left + right) >> 1);
if(target == nums[mid])
{
res = true;
break;
}
else if(nums[mid] > nums[left]) //前半部分有序
{
if(target >= nums[left] && target < nums[mid])
right = mid - 1;
else
left = mid + 1;
}
else if(nums[mid] < nums[left]) //后半部分有序
{
if(target > nums[mid] && target <= nums[right])
left = mid + 1;
else
right = mid - 1;
}
else //nums[mid]==nums[left]的情况，包括mid=left和存在重复值两种情况
left++;
}
return res;
}
};

nums[mid]与nums[right]比较的代码：

class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
bool res = false;
int left = 0, right = n - 1;

while(left <= right)
{
int mid = (left + right) >> 1;
if(target == nums[mid])
{
res = true;
break;
}
else if(nums[mid] < nums[right]) //后半部分有序
{
if(target > nums[mid] && target <= nums[right])
left = mid + 1;
else
right = mid - 1;
}
else if(nums[mid] > nums[right]) //前半部分有序
{
if(target >= nums[left] && target < nums[mid])
right = mid - 1;
else
left = mid + 1;
}
else //存在重复值的情况
right--;
}
return res;
}
};