备忘
非常好的题目,虽然很简单但是是对二分查找思想的经典考察。
y总提供的二分查找的两种模板正好对应题目的两种情况,这道题需要牢记。
C++ 代码
class Solution {
public:
int search(vector<int>& nums, int target) {
if(nums.empty())return 0;
int left = bisearch2(nums, target, 0, nums.size() - 1);
if(nums[left] != target)
return 0;
int right = bisearch1(nums, target, 0, nums.size() - 1);
return right - left + 1;
}
int bisearch1(vector<int>& nums, int target, int l, int r){
if(l == r)return l;
while(l < r){
int mid = (l + r + 1) / 2;
if(nums[mid] <= target)
l = mid;
else
r = mid - 1;
}
return l;
}
int bisearch2(vector<int>& nums, int target, int l, int r){
if(l == r)return l;
while(l < r){
int mid = (l + r) / 2;
if(nums[mid] >= target)
r = mid;
else
l = mid + 1;
}
return l;
}