给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
你的算法时间复杂度必须是 O(log n) 级别。
如果数组中不存在目标值,返回 [-1, -1]。
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]
明显是二分法的题,既然有两个边界,那就找两次,第一次找左边,第二次找右边,要注意的一点是,由于计算中间值时除法是向下取整的,所以只能从左边向右靠近,所以都是left+1的情况,其余的一些细节直接见程序即可。
C++源代码:
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res(2,-1);
if(nums.size()==0) return res;
int left = 0, right = nums.size()-1;
while(left < right)
{
int mid = left + (right-left) / 2;
if(nums[mid] < target)
left = mid + 1;
else
right = mid;
}
if(target!=nums[left])
return res;
res[0] = left;
right = nums.size();
while(left < right)
{
int mid = left + (right-left) / 2;
if(nums[mid] <= target)
left = mid+1;
else
right = mid;
}
res[1] = right-1;
return res;
}
};
python3源代码:
class Solution:
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
res = [-1, -1]
if len(nums)==0: return res
left = 0
right = len(nums)-1
while left < right:
mid = left + (right-left) // 2
if nums[mid] < target:
left = mid+1
else:
right = mid
if nums[left]!=target: return res
res[0] = left
right = len(nums)
while left < right:
mid = left + (right-left) // 2
if nums[mid] <= target:
left = mid + 1
else:
right = mid
res[1] = right - 1
return res