准备每天没事刷刷 leetcode 并用这个 markdown 文档做个简单记录
c++:
class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
vector<int> res;
for (int i = 0; i < nums.size(); i++)
res.push_back(compare_nums(nums, nums[i]));
return res;
}
int compare_nums(vector<int>& nums, int temp) {
int res = 0;
for (int i = 0; i < nums.size(); i++)
if (nums[i] < temp)
res++;
return res;
}
};
python:
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
res = []
for c in nums:
res.append(len([1 for i in nums if i < c]))
return respython:
class Solution:
def numberOfSteps(self, num: int) -> int:
res = 0
while num != 0:
num = num // 2 if num % 2 == 0 else num - 1
res += 1
return resc++:
class Solution {
public:
int numberOfSteps(int num) {
int res = 0;
while (num != 0) {
num = (num % 2 == 0)? (num / 2) :( num-1);
res++;
}
return res;
}
};
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special solvement:
Count each 0 as 1. Count each 1 as 2 except the first 1. Time: O(bits of num) Space: O(bits of num) However num is bounded in 0 <= num <= 10^6, so time and space are both O(1) in this problem.
class Solution:
def numberOfSteps (self, num: int) -> int:
digits = f'{num:b}'
return digits.count('1') - 1 + len(digits)O(1) space:
class Solution:
def numberOfSteps (self, num: int) -> int:
steps = num == 0
while num > 0:
steps += 1 + (num & 1)
num >>= 1
return steps - 1python:
class Solution:
def subtractProductAndSum(self, n: int) -> int:
str_n = str(n)
product = 1
sums = 0
for c in str_n:
product *= int(c)
sums += int(c)
return product - sumsone-line solution:
def subtractProductAndSum(n):
return eval('*'.join(str(n))) - eval('+'.join(str(n)))other solutions:
import numpy as np
class Solution:
def subtractProductAndSum(self, n: int) -> int:
a = [int(x) for x in str(n)]
return np.prod(a) - np.sum(a)from functools import reduce
class Solution:
def subtractProductAndSum(self, n: int) -> int:
a = [int(x) for x in str(n)]
return reduce((lambda x, y: x * y), a) - reduce((lambda x, y: x + y), a)faster methods:
def subtractProductAndSum(self, n):
dm = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
s = repr(n)
sm, pr = 0, 1
for c in s:
pr *= dm[c]
sm += dm[c]
return pr-smusing map methods to solve the list data type
def subtractProductAndSum(self, n):
A = map(int, str(n))
return reduce(operator.mul, A) - sum(A)
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