mirror of
https://github.com/SunnyQjm/algorithm-homework.git
synced 2026-10-05 01:15:36 +08:00
111 lines
3.6 KiB
Markdown
111 lines
3.6 KiB
Markdown
# Algorithm Homework 2
|
||
|
||
> 姓名：阙建明
|
||
>
|
||
> 学号：1901213139
|
||
|
||
## 作业要求：
|
||
|
||
|
||
|
||
## 题解
|
||
|
||
- ### Leetcode problem 7（[整数反转](https://leetcode-cn.com/problems/reverse-integer/)）
|
||
|
||
- **算法思路：**
|
||
|
||
- 首先记录是否为负数，接着将输入值取整
|
||
- 用模10取余的方式依次得到反转的每一位，然后拼接到一个 *result* 当中
|
||
- 最后判断是否溢出，溢出则返回0，不溢出则返回 *result* 或 *-result*
|
||
|
||
- **Leetcode提交结果：**
|
||
|
||
|
||
|
||
- **代码：**
|
||
|
||
```python
|
||
INT_MAX_VAL = 2147483647
|
||
INT_MIN_VAL = -2147483648
|
||
class Solution:
|
||
def reverse(self, x: int) -> int:
|
||
isNegitive = -1 if x < 0 else 1
|
||
result, x = 0, abs(x)
|
||
while x > 0:
|
||
result = result * 10 + x % 10
|
||
x //= 10
|
||
return 0 if result > INT_MAX_VAL or result < INT_MIN_VAL else isNegitive * result
|
||
```
|
||
|
||
- ### Leetcode problem 13（[罗马数字转整数](https://leetcode-cn.com/problems/roman-to-integer/)）
|
||
|
||
- **算法思路：**
|
||
|
||
- 首先对于 *M、D、L、V*，可直接返回对应代表的数值
|
||
- 对于 *C、X、I*，则如果右边出现可与之搭配成9 × 10n 或者 4 × 10n（n = 0, 1, 2）的情况，返回负的对应数值，否则返回正的对应数值
|
||
|
||
- **Leetcode提交结果：**
|
||
|
||
|
||
|
||
- **代码：**
|
||
|
||
```python
|
||
class Solution:
|
||
def romanToInt(self, s: str) -> int:
|
||
result, length = 0, len(s)
|
||
for i in range(0, length):
|
||
if s[i] == 'M':
|
||
result += 1000
|
||
elif s[i] == 'D':
|
||
result += 500
|
||
elif s[i] == 'C':
|
||
if i < length - 1 and (s[i + 1] == 'M' or s[i + 1] == 'D'):
|
||
result -= 100
|
||
else:
|
||
result += 100
|
||
elif s[i] == 'L':
|
||
result += 50
|
||
elif s[i] == 'X':
|
||
if i < length - 1 and (s[i + 1] == 'L' or s[i + 1] == 'C'):
|
||
result -= 10
|
||
else:
|
||
result += 10
|
||
elif s[i] == 'V':
|
||
result += 5
|
||
elif s[i] == 'I':
|
||
if i < length - 1 and (s[i + 1] == 'V' or s[i + 1] == 'X'):
|
||
result -= 1
|
||
else:
|
||
result += 1
|
||
return result
|
||
```
|
||
|
||
- ### Leetcode problem 3（[加一](https://leetcode-cn.com/problems/plus-one/)）
|
||
|
||
- **算法思路：**
|
||
|
||
- 从后往前遍历，每次 + 1，如果和等于10，则发生进位，否则算法返回
|
||
- 遍历结束发现 *tmp* 仍然不为0，则表示数组的值为全9，需要在数组前添加一个元素
|
||
|
||
- **Leetcode提交结果：**
|
||
|
||
|
||
|
||
- **代码：**
|
||
|
||
```python
|
||
class Solution:
|
||
def plusOne(self, digits: List[int]) -> List[int]:
|
||
tmp = digits[len(digits) - 1] + 1
|
||
digits[len(digits) - 1], tmp, index = tmp % 10, tmp // 10, len(digits) - 2
|
||
while tmp != 0 and index >= 0:
|
||
tmp = digits[index] + 1
|
||
digits[index], tmp = tmp % 10, tmp // 10
|
||
index -= 1
|
||
if tmp != 0:
|
||
digits.insert(0, 1)
|
||
return digits
|
||
```
|
||
|
||
|