33. 81. Search in Rotated Sorted Array *HARD*
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.
利用二分查找的思想。
int binary_search(vector<int>& nums, int left, int right, int target)
{
if(left > right)
return -;
while(left <= right)
{
int mid = (left+right)>>;
if(nums[mid] == target)
return mid;
else if(nums[mid] < target)
left = mid+;
else
right = mid-;
}
return -;
}
int search(vector<int>& nums, int target) {
int n = nums.size(), left = , right = n-, mid;
if(left == right && nums[] == target)
return ;
while(left <= right)
{
mid = (left+right)>>;
if(target == nums[mid])
return mid;
if(nums[left] <= nums[mid])
{
if(target >= nums[left] && target <= nums[mid])
return binary_search(nums, left, mid, target);
left = mid+;
}
else
{
if(target >= nums[mid] && target <= nums[right])
return binary_search(nums, mid, right, target);
right = mid-;
}
}
return -;
}
What if duplicates are allowed?
class Solution {
public:
bool search(vector<int>& nums, int target) {
int n = nums.size(), left = , right = n-, mid;
bool found = false;
while(left <= right)
{
mid = (left+right) >> ;
if(nums[mid] == target)
return true;
if(nums[left] < nums[mid])
{
if(target >= nums[left] && target < nums[mid])
right = mid - ;
else
left = mid + ;
}
else if(nums[left] > nums[mid])
{
if(target > nums[mid] && target <= nums[right])
left = mid + ;
else
right = mid - ;
}
else
left++;
}
return false;
}
};
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