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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。 + + - **Leetcode提交结果：** + + ![image-20200407102849523](AlgorithmHomework7.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： + + $P(i, j) = \begin{cases}P(i, j) + P(i + 1, j) &\text{if } P(i, j + 1) >= P(j + 1) \\P(i, j) + P(i, j + 1) &\text{if }P(i, j + 1) < P(j + 1) \end{cases}$ + + - **Leetcode提交结果：** + + ![image-20200407105454180](AlgorithmHomework7.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： + + $P(row, index) = \begin{cases}P(row, index) &\text{if } row == len(triangle) - 1 \\P(row, index) + min(P(row + 1, index), P(row + 1, index + 1)) &\text{if }row < len(triangle) - 1 \end{cases}$ + + - **Leetcode提交结果：** + + ![image-20200407122505839](AlgorithmHomework7.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] + ``` \ No newline at end of file diff --git a/README.md b/README.md old mode 100644 new mode 100755 diff --git a/index.html b/index.html old mode 100644 new mode 100755 index 9995db3..a9f48e7 --- a/index.html +++ b/index.html @@ -1,103 +1,103 @@ - - - - - 算法课作业-阙建明 - - - - - - - - - - -
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