algorithm-homework/AlgorithmHomework6.md

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

  • 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：

    

  • 代码：

    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

  • 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：

    

  • 代码：

    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

  • 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：

    

  • 代码：

    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]