算法1
思路:使用两种算法模版的差异,首先找出大于等于target的最小的数的位置,然后找出小于等于target的最大的数的位置。
时间复杂度
2O(logn)
C++ 代码
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
if (nums.empty()) return vector<int>({-1, -1});
vector<int> res;
int l = 0, r = nums.size() - 1;
while(l < r)
{
int mid = l + r >> 1;
if(nums[mid] >= target) r = mid;
else l = mid+ 1;
}
if (nums[l] != target) return vector<int>({-1, -1});
res.push_back(l);
l = 0, r = nums.size() - 1;
while(l < r)
{
int mid = l + r + 1 >> 1;
if(nums[mid] <= target) l = mid;
else r = mid - 1;
}
res.push_back(l);
return res;
}
};