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2026-10-05 01:15:36 +08:00 · 2020-03-20 13:45:13 +08:00
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+# Algorithm Homework 1
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+- ## 作业要求
+  3道上课讲的LeetCode题目1，69，70
+
+- ## Leetcode problem 1 (两数之和)
+  - 算法思路：<br/>
+    用字典记录数字，以便在找到成对的另一个数的时候，可以快速的检索到前一个数的索引
+  
+  - Leetcode提交结果
+    ![result](AlgorithmHomework1.assets/7222676-c485f42b177afce2.png)
+  - 代码：
+    ```python
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        myDict = {}
+        for index, num in enumerate(nums):
+            if num in myDict:
+                return [myDict[num], index]
+            else:
+                myDict[target - num] = index
+    ```
+
+- ## Leetcode problem 69 (x 的平方根)
+  - 算法思路：二分法
+  
+  - Leetcode提交结果
+    ![result](AlgorithmHomework1.assets/7222676-6a5936bab1f54a06.png)
+  - 代码：
+    ```python
+    def mySqrt(self, x: int) -> int:
+        if x in [1, 2, 3]:
+            return 1
+        left = 0
+        right = x // 2
+        while True:
+            mid = (left + right) // 2
+            if left + 1 >= right:
+                break
+            if mid < x / mid:  # mid to small
+                left = mid
+            else:
+                right = mid
+        return right if right * right <= x else left
+    ```
+
+- ## Leetcode problem 70 (爬楼梯)
+  - 算法思路：动态规划。由于当前状态只取决于前两个状态，所以dp table可以精简为只用两个数表示
+  
+  - Leetcode提交结果
+    ![result](AlgorithmHomework1.assets/7222676-dde7e98b91cb5454.png)
+  - 代码：
+    ```python
+    def climbStairs(self, n: int) -> int:
+        pre, cur = 0, 1
+        for i in range(1, n + 1):
+            pre, cur = cur, pre + cur
+        return cur
+    ```
@@ -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
+    ```
+
+    
@@ -0,0 +1,111 @@
+# Algorithm Homework 3
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+Leetcode: 35,58,83
+
+## 题解：
+
+- ### Leetcode problem 35（[搜索插入位置](https://leetcode-cn.com/problems/search-insert-position/)）
+
+  - **算法思路：** 
+
+    - 尝试使用二分法找到target的索引，找到则返回
+    - 没有找到则判断target与左右索引的大小关系，返回应该插入的位置
+
+  - **Leetcode提交结果：**
+
+    ![image-20200313210334196](https://raw.githubusercontent.com/SunnyQjm/picgo_0/master/img/image-20200313210334196.png)
+
+  - **代码：**
+
+    ```python
+    class Solution:
+        def searchInsert(self, nums: List[int], target: int) -> int:
+            left, right = 0, len(nums) - 1
+            while left < right - 1:
+                pos = (left + right) // 2
+                if nums[pos] > target:
+                    right = pos
+                elif nums[pos] < target:
+                    left = pos
+                else:
+                    return pos
+            if nums[left] == target or target < nums[left]:
+                return left
+            elif nums[right] == target or target < nums[right]:
+                return right
+            else:
+                return right + 1
+    ```
+
+- ### Leetcode problem 58（[最后一个单词的长度](https://leetcode-cn.com/problems/length-of-last-word/)）
+
+  - **算法思路：** 
+
+    - 思路1：使用`split`将字符串用空格分割成数组，然后反序遍历，取第一个非空字符串的长度，无则返回0；
+    - 思路2：使用`strip`截掉字符串前后的空格，然后反序遍历统计最后一个单次的长度，无则返回0。
+
+  - **Leetcode提交结果：**
+
+    ![image-20200313211905775](https://raw.githubusercontent.com/SunnyQjm/picgo_0/master/img/image-20200313213736914.png)
+
+  - **代码：**
+
+    ```python
+    # 思路1代码
+    class Solution:
+        def lengthOfLastWord(self, s: str) -> int:
+            strs = s.split(' ')
+            for i in range(0, len(strs))[::-1]:
+                if len(strs[i]) > 0: return len(strs[i])
+            return 0
+    
+    # 思路2代码（性能更优）
+    class Solution:
+        def lengthOfLastWord(self, s: str) -> int:
+            s = s.strip()
+            count = 0
+            for i in range(0, len(s))[::-1]:
+                if s[i] == ' ':
+                    return count
+                else:
+                    count += 1
+            return count
+    ```
+
+- ### Leetcode problem 83（[删除排序链表中的重复元素](https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/)）
+
+  - **算法思路：** 
+
+    - 用一个set记录当前遍历过的值
+    - 遍历到每一个节点，用set检查当前值是否存在于set，如果存在则删除该节点
+
+  - **Leetcode提交结果：**
+
+    ![image-20200313213736914](https://raw.githubusercontent.com/SunnyQjm/picgo_0/master/img/image-20200313211905775.png)
+
+  - **代码：**
+
+    ```python
+    class Solution:
+        def deleteDuplicates(self, head: ListNode) -> ListNode:
+            if head == None:
+                return head
+            s, pre, cur = set(), head, head.next
+            s.add(pre.val)
+            while cur is not None:
+                if cur.val in s:
+                    pre.next, cur = cur.next, cur.next
+                    continue
+                else:
+                    s.add(cur.val)
+                pre, cur = cur, cur.next
+            return head
+    ```
+
+    
@@ -1 +1,5 @@
 # algorithm-homework
+
+- [第一次作业](AlgorithmHomework1.md)
+- [第二次作业](AlgorithmHomework2.md)
+- [第三次作业](AlgorithmHomework3.md)
@@ -0,0 +1,103 @@
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