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+# Algorithm Homework 6
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+Leetcode: 5, 64, 120
+
+## 题解：
+
+- ### Leetcode problem 5
+
+  >   [最长回文子串](https://leetcode-cn.com/problems/longest-palindromic-substring/)
+
+  - **算法思路：** 
+
+    - 使用动态规划的思想；
+
+    - 状态转移方程如下：
+      $$
+      P(i, j) = \left\{ \begin{array}{ll}
+      True & \textrm{if $P(i + 1, j - 1) $ and $ S_i == S_j $}\\
+      False & \textrm{其它}\\
+      \end{array} \right.
+      $$
+
+    - 首先判断长度为1的子串，再判断长度为2的子串，再判断长度为3的子串，以此类推。
+
+  - **Leetcode提交结果：**
+
+    ![image-20200407102849523](AlgorithmHomework6.assets/image-20200407102849523.png)
+
+  - **代码：**
+
+    ```python
+    class Solution:
+        def longestPalindrome(self, s: str) -> str:
+            arr = [[False] * len(s) for i in range(len(s))]
+            for i in range(len(s)):
+                arr[i][i] = True
+            maxLen, maxStrStart, maxStrEnd = 1, 0, 1
+            for span in range(1, len(s)):
+                left, right = 0, span
+                while right < len(s):
+                    arr[left][right] = True if s[left] == s[right] and (span == 1 or arr[left + 1][right - 1]) else False
+                    if arr[left][right] and right - left + 1 > maxLen:
+                        maxLen, maxStrStart, maxStrEnd = right - left + 1, left, right + 1
+                    left, right = left + 1, right + 1
+            return s[maxStrStart: maxStrEnd]
+    ```
+
+- ### Leetcode problem 64
+
+  >   [最小路径和](https://leetcode-cn.com/problems/minimum-path-sum/)
+
+  - **算法思路：** 
+
+    - 采用动态规划的思想，自右下角向左上脚遍历。
+
+    - 首先将最下面一排的路径开销自右往左以此叠加，把最右边一排的路径开销自下往上以此叠加，接着其它情况使用与下述状态转换方程：
+
+    $$
+    P(i, j) = \left\{ \begin{array}{ll}
+            P(i, j) + P(i + 1, j) & \textrm{if $ P(i, j + 1) >= P(j + 1) $}\\
+          P(i, j) + P(i, j + 1) & \textrm{if $ P(i, j + 1) < P(j + 1) $}\\
+            \end{array} \right.
+    $$
+
+  - **Leetcode提交结果：**
+
+    ![image-20200407105454180](AlgorithmHomework6.assets/image-20200407105454180.png)
+
+  - **代码：**
+
+    ```python
+    class Solution:
+        def minPathSum(self, grid: List[List[int]]) -> int:
+            rowNum, colNum = len(grid), len(grid[0])
+            for col in range(colNum - 2, -1, -1):
+                grid[rowNum - 1][col] = grid[rowNum - 1][col] + grid[rowNum - 1][col + 1]
+            for row in range(rowNum - 2, -1, -1):
+                grid[row][colNum - 1] = grid[row][colNum - 1] + grid[row + 1][colNum - 1]
+            for col in range(colNum - 2, -1, -1):
+                for row in range(rowNum - 2, -1, -1):
+                    grid[row][col] = grid[row][col] + min(grid[row + 1][col], grid[row][col + 1])
+            return grid[0][0]
+    ```
+
+- ### Leetcode problem 120
+
+  >  [三角形最小路径和](https://leetcode-cn.com/problems/triangle/)
+
+  - **算法思路：** 
+
+    - 用动态规划的思想，自底向上一次计算到当前访问节点到叶子节点的最短路径；
+  
+    - 状态转移方程如下：
+  
+    $$
+    P(row, index) = \left\{ \begin{array}{ll}
+                  P(row, index) & \textrm{if $ row == len(triangle) - 1 $}\\
+              P(row, index) + min(P(row + 1, index), P(row + 1, index + 1))& \textrm{if $ P(i, j + 1) < P(j + 1) $}\\
+                  \end{array} \right.
+    $$
+  
+  - **Leetcode提交结果：**
+  
+    ![image-20200407122505839](AlgorithmHomework6.assets/image-20200407122505839.png)
+  
+  - **代码：**
+  
+    ```python
+    class Solution:
+      def minimumTotal(self, triangle: List[List[int]]) -> int:
+            for row in range(len(triangle) - 2, -1, -1):
+                for index in range(len(triangle[row])):
+                    triangle[row][index] = triangle[row][index] + min(triangle[row + 1][index], triangle[row + 1][index + 1])
+            return triangle[0][0]
+    ```
+    
+  
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