题目:统计一个数字在排序数组中出现的次数。
解法一:
思路:暴力解法。遍历一遍数组,统计该数字出现的次数。
class Solution {
public:
int GetNumberOfK(vector<int> data ,int k) {
int count = 0;
for (int i= 0;i<data.size();++i)
{
if(data[i]==k)
count++;
}
return count;
}
};
解法二:
思路:两次二分法,分别确定数字出现的首尾。作差即为答案。
class Solution {
public:
int GetNumberOfK(vector<int> data, int k) {
int l = 0;
int r = data.size();
while (l < r)
{
int mid = (l + r) / 2;
if (data[mid] < k) l = mid + 1;
else
r = mid;
}
int left = l;
l = 0;
r = data.size();
while (l < r)
{
int mid = (l + r) / 2;
if (data[mid] > k) r=mid;
else
l = mid+1;
}
return r-left;
}
};
python的解法:
暴力法:
# -*- coding:utf-8 -*-
class Solution:
def GetNumberOfK(self, data, k):
# write code here
count = 0
for num in data:
if num == k:
count += 1
return count