文章目录
一，路径交叉
1，程序简介
给你一个整数数组 distance 。
从 X-Y 平面上的点 (0,0) 开始，先向北移动 distance[0] 米，然后向西移动 distance[1] 米，向南移动 distance[2] 米，向东移动 distance[3] 米，持续移动。也就是说，每次移动后你的方位会发生逆时针变化。
判断你所经过的路径是否相交。如果相交，返回 true ；否则，返回 false 。
示例 1：
- 输入：distance = [2,1,1,2]
- 输出：true
示例 2：
- 输入：distance = [1,2,3,4]
- 输出：false
示例 3：
- 输入：distance = [1,1,1,1]
- 输出：true
提示：
- 1 < = d i s t a n c e . l e n g t h < = 1 0 5 1 <= distance.length <= 10^5 1<=distance.length<=105
- 1 < = d i s t a n c e [ i ] < = 1 0 5 1 <= distance[i] <= 10^5 1<=distance[i]<=105
2，程序代码
# -*- coding: utf-8 -*-
"""
Created on Sat Feb 19 06:23:44 2022
Function: 路径交叉
@author: 小梁aixj
"""
class Solution:
def isSelfCrossing(self, x: List[int]) -> bool:
n = len(x)
if n < 3: return False
for i in range(3, n):
if x[i] >= x[i - 2] and x[i - 3] >= x[i - 1]:
return True
if i >= 4 and x[i - 3] == x[i - 1] and x[i] + x[i - 4] >= x[i - 2]:
return True
if i >= 5 and x[i - 3] >= x[i - 1] and x[i - 2] >= x[i - 4] and x[i - 1] + x[i - 5] >= x[i - 3] and x[i] + x[i - 4] >= x[i - 2]:
return True
return False
二，阶乘后的零
1，程序简介
给定一个整数 n ，返回 n! 结果中尾随零的数量。
提示 n! = n * (n - 1) * (n - 2) * … * 3 * 2 * 1
示例 1：
- 输入：n = 3
- 输出：0
- 解释：3! = 6 ，不含尾随 0
示例 2：
- 输入：n = 5
- 输出：1
- 解释：5! = 120 ，有一个尾随 0
示例 3：
- 输入：n = 0
- 输出：0
提示：
- 0 < = n < = 1 0 4 0 <= n <= 10^4 0<=n<=104
进阶：
- 你可以设计并实现对数时间复杂度的算法来解决此问题吗？
2，程序代码
# -*- coding: utf-8 -*-
"""
Created on Sat Feb 19 06:25:53 2022
Function: 阶乘后的零
@author: 小梁aixj
"""
class Solution:
def trailingZeroes(self, n: int) -> int:
zero_count = 0
current_multiple = 5
while n >= current_multiple:
zero_count += n // current_multiple
current_multiple *= 5
return zero_count
三，整数转换英文表示
1，程序简介
将非负整数 num 转换为其对应的英文表示。
示例 1：
- 输入：num = 123
- 输出：“One Hundred Twenty Three”
示例 2：
- 输入：num = 12345
- 输出：“Twelve Thousand Three Hundred Forty Five”
示例 3：
- 输入：num = 1234567
- 输出：“One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven”
示例 4：
- 输入：num = 1234567891
- 输出：“One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One”
提示：
- 0 < = n u m < = 2 31 − 1 0 <= num <= 2^{31} - 1 0<=num<=231−1
2，程序代码
# -*- coding: utf-8 -*-
"""
Created on Sat Feb 19 06:38:47 2022
Function: 整数转换英文表示
@author: 小梁aixj
"""
class Solution(object):
def numberToWords(self, num):
"""
:type num: int
:rtype: str
"""
def helper(num):
n = int(num)
num = str(n)
if n < 100:
return subhelper(num)
else:
return ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"][int(num[0]) - 1] + " Hundred " + subhelper(num[1:]) if num[1:] != "00" else ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"][int(num[0]) - 1] + " Hundred"
def subhelper(num):
n = int(num)
l1 = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"]
l2 = ["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"]
l3 = ["Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]
if n < 10:
return l1[int(num)]
if 10 <= n < 20:
return l2[n - 10]
if 20 <= n < 100:
return l3[int(num[0]) - 2] + " " + l1[int(num[1])] if num[1] != "0" else l3[int(num[0]) - 2]
res = ""
if num >= 1000000000:
res = helper(str(num)[0]) + " Billion"
if str(num)[1:4] != "000":
res += " " + helper(str(num)[1:4]) + " Million"
if str(num)[4:7] != "000":
res += " " + helper(str(num)[4:7]) + " Thousand"
if str(num)[7:] != "000":
res += " " + helper(str(num)[7:])
elif num >= 1000000:
res = helper(str(num)[:-6]) + " Million"
if str(num)[-6:-3] != "000":
res += " " + helper(str(num)[-6:-3]) + " Thousand"
if str(num)[-3:] != "000":
res += " " + helper(str(num)[-3:])
elif num >= 1000:
res = helper(str(num)[:-3]) + " Thousand"
if str(num)[-3:] != "000":
res += " " + helper(str(num)[-3:])
else:
return helper(str(num))
return res
