update

AlgorithmHomework2.md

@@ -0,0 +1,111 @@
+# Algorithm Homework 2
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+![image-20200320133934742](AlgorithmHomework2.assets/image-20200320133934742.png)
+
+## 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：**
+
+    ![image-20200320133949475](AlgorithmHomework2.assets/image-20200320133949475.png)
+
+  - **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：**
+
+    ![image-20200320134000771](AlgorithmHomework2.assets/image-20200320134000771.png)
+
+  - **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：**
+
+    ![image-20200320134010709](AlgorithmHomework2.assets/image-20200320134010709.png)
+
+  - **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
+    ```
+
+