add: homework8

AlgorithmHomework3.md

@@ -21,7 +21,7 @@ Leetcode: 35,58,83
 
   - **Leetcode提交结果：**
 
-    ![image-20200313210334196](https://raw.githubusercontent.com/SunnyQjm/picgo_0/master/img/image-20200313210334196.png)
+    ![image-20200313210334196](AlgorithmHomework3.assets/image-20200313210334196.png)
 
   - **代码：**
 
@@ -56,7 +56,7 @@ Leetcode: 35,58,83
 
   - **Leetcode提交结果：**
 
-    ![image-20200313211905775](https://raw.githubusercontent.com/SunnyQjm/picgo_0/master/img/image-20200313213736914.png)
+    ![image-20200313211905775](AlgorithmHomework3.assets/image-20200313213736914.png)
 
   - **代码：**
 
@@ -89,12 +89,13 @@ Leetcode: 35,58,83
   - **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)
-
+    
+- 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](AlgorithmHomework3.assets/image-20200313211905775.png)
+  
   - **代码：**
   
     ```python

AlgorithmHomework8.md

@@ -0,0 +1,146 @@
+# Algorithm Homework 8
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+Leetcode: 72, 312, 729
+
+## 题解：
+
+- ### Leetcode problem 72
+
+  > [编辑距离](https://leetcode-cn.com/problems/edit-distance/)
+
+  - **算法思路：** 
+      1. `dp[i][j] `表示word1的前i个字母和word2的前j个字母的编辑距离
+      2. 在word1中删除一个字符，等价于在word2中适当位置添加一个字符
+      3. 如果`word1[i - 1]` == `word2[j - 1]`:
+          - `dp[i - 1][j]`的基础上在word1上增加一个字符 => 操作+1
+          - `dp[i][j - 1]`的基础上在word2上增加一个字符 => 操作+1
+          - `dp[i - 1][j - 1]`的基础上不做操作 => 操作+0
+      3. 如果word1[i - 1] != word2[j - 1]
+          - `dp[i - 1][j]`的基础上在word1上增加一个字符 => 操作+1
+          - `dp[i][j - 1]`的基础上在word2上增加一个字符 => 操作+1
+          - `dp[i - 1][j - 1]`的基础上修改`word1[i]`，使得`word1[i] = word2[j]` => 操作+1
+
+  - **Leetcode提交结果：**
+
+    ![image-20200423115932875](AlgorithmHomework8.assets/image-20200423115932875.png)
+
+  - **代码：**
+
+    ```python
+    class Solution(object):
+        def minDistance(self, word1, word2):
+            """
+            :type word1: str
+            :type word2: str
+            :rtype: int
+            """
+            n, m = len(word1), len(word2)
+            if n * m == 0:
+                return n + m
+            dp = [[0] * (m + 1) for _ in range(n + 1)]
+            for i in range(n + 1):
+                dp[i][0] = i
+            for j in range(m + 1):
+                dp[0][j] = j
+    
+            for i in range(1, n + 1):
+                for j in range(1, m + 1):
+                    tmp1 = dp[i - 1][j] + 1
+                    tmp2 = dp[i][j - 1] + 1
+                    tmp3 = dp[i - 1][j - 1]
+                    if word1[i - 1] != word2[j - 1]:
+                        tmp3 += 1
+                    dp[i][j] = min(tmp1, tmp2, tmp3)
+            return dp[-1][-1]
+    ```
+
+- ### Leetcode problem 312
+
+  > [ 戳气球](https://leetcode-cn.com/problems/burst-balloons/)
+
+  - **算法思路：** 
+
+    - 采用动态规划的思想，`dp[left][right]` 表示戳破开区间（left, right）中的气球所能得到的最大金币数量；
+  
+  - 为了处理边缘的数据，这边在初始数组的左右两边各加一个值为1的元素；
+  
+  - 考虑从没有气球开始把气球逐个加入，则添加第i个气球后能得到的最大金币数为：
+  
+    $nums[left] * nums[i] * nums[right] + dp(left, i) + dp(i, right)$
+  
+- **Leetcode提交结果：**
+  
+    ![image-20200423211525664](AlgorithmHomework8.assets/image-20200423211525664.png)
+  
+  - **代码：**
+  
+    ```python
+    class Solution(object):
+        def maxCoins(self, nums):
+            """
+            :type nums: List[int]
+            :rtype: int
+            """
+            nums = [1] + nums + [1]
+            n = len(nums)
+            dp = [[0] * n for _ in range(n)]
+            for left in range(n-2, -1, -1):
+                for right in range(left+2, n):
+                    dp[left][right] = max(nums[left] * nums[i] * nums[right] + dp[left][i] + dp[i][right] for i in range(left+1, right))
+            return dp[0][n-1]
+    ```
+  
+- ### Leetcode problem 729
+
+  > [我的日程安排表 I](https://leetcode-cn.com/problems/my-calendar-i/)
+
+  - **算法思路：** 
+
+    - 每次判断现有的行程中有没有和当前要添加的行程有重叠的部分；
+  - 如果有重叠，则直接返回Fase
+    - 如果没有重叠，则将新的行程插入到特定位置，保证行程数组是按序插入的
+
+  - **Leetcode提交结果：**
+
+    ![image-20200423221735514](AlgorithmHomework8.assets/image-20200423221735514.png)
+
+  - **代码：**
+
+    ```python
+    class MyCalendar(object):
+    
+        def __init__(self):
+            self.calendar = []
+    
+        def book(self, start, end):
+            """
+            :type start: int
+            :type end: int
+            :rtype: bool
+            """
+    
+            if len(self.calendar) == 0:
+                self.calendar.append([start, end])
+                return True
+
+            for i in range(len(self.calendar)):
+                if self.calendar[i][0] >= end:
+                    if i == 0:
+                        self.calendar.insert(0, [start, end])
+                        return True
+                    elif self.calendar[i - 1][1] <= start:
+                        self.calendar.insert(i, [start, end])
+                        return True
+            if self.calendar[-1][1] <= start:
+                self.calendar.append([start, end])
+                return True
+            return False
+    ```
+    
+