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108 lines
3.8 KiB
Markdown
Executable File
108 lines
3.8 KiB
Markdown
Executable File
# Algorithm Homework 6
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> 姓名：阙建明
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>
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> 学号：1901213139
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## 作业要求：
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Leetcode: 5, 64, 120
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## 题解：
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- ### Leetcode problem 5
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> [最长回文子串](https://leetcode-cn.com/problems/longest-palindromic-substring/)
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- **算法思路：**
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- 使用动态规划的思想；
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- 状态转移方程如下：
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$P(i, j) = \begin{cases}True &\text{if } P(i + 1, j - 1)==True , S_i == S_j \\False &\text{if 其它 } \end{cases}$
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- 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。
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- **Leetcode提交结果：**
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- **代码：**
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```python
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class Solution:
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def longestPalindrome(self, s: str) -> str:
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arr = [[False] * len(s) for i in range(len(s))]
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for i in range(len(s)):
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arr[i][i] = True
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maxLen, maxStrStart, maxStrEnd = 1, 0, 1
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for span in range(1, len(s)):
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left, right = 0, span
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while right < len(s):
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arr[left][right] = True if s[left] == s[right] and (span == 1 or arr[left + 1][right - 1]) else False
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if arr[left][right] and right - left + 1 > maxLen:
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maxLen, maxStrStart, maxStrEnd = right - left + 1, left, right + 1
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left, right = left + 1, right + 1
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return s[maxStrStart: maxStrEnd]
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```
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- ### Leetcode problem 64
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> [最小路径和](https://leetcode-cn.com/problems/minimum-path-sum/)
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- **算法思路：**
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- 采用动态规划的思想，自右下角向左上脚遍历。
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- 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：
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$P(i, j) = \begin{cases}P(i, j) + P(i + 1, j) &\text{if } P(i, j + 1) >= P(j + 1) \\P(i, j) + P(i, j + 1) &\text{if }P(i, j + 1) < P(j + 1) \end{cases}$
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- **Leetcode提交结果：**
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- **代码：**
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```python
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class Solution:
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def minPathSum(self, grid: List[List[int]]) -> int:
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rowNum, colNum = len(grid), len(grid[0])
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for col in range(colNum - 2, -1, -1):
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grid[rowNum - 1][col] = grid[rowNum - 1][col] + grid[rowNum - 1][col + 1]
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for row in range(rowNum - 2, -1, -1):
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grid[row][colNum - 1] = grid[row][colNum - 1] + grid[row + 1][colNum - 1]
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for col in range(colNum - 2, -1, -1):
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for row in range(rowNum - 2, -1, -1):
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grid[row][col] = grid[row][col] + min(grid[row + 1][col], grid[row][col + 1])
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return grid[0][0]
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```
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- ### Leetcode problem 120
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> [三角形最小路径和](https://leetcode-cn.com/problems/triangle/)
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- **算法思路：**
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- 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；
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- 状态转移方程如下：
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$P(row, index) = \begin{cases}P(row, index) &\text{if } row == len(triangle) - 1 \\P(row, index) + min(P(row + 1, index), P(row + 1, index + 1)) &\text{if }row < len(triangle) - 1 \end{cases}$
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- **Leetcode提交结果：**
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- **代码：**
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```python
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class Solution:
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def minimumTotal(self, triangle: List[List[int]]) -> int:
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for row in range(len(triangle) - 2, -1, -1):
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for index in range(len(triangle[row])):
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triangle[row][index] = triangle[row][index] + min(triangle[row + 1][index], triangle[row + 1][index + 1])
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return triangle[0][0]
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```
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