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add: homework7
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- 算法思路：
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用字典记录数字，以便在找到成对的另一个数的时候，可以快速的检索到前一个数的索引
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- Leetcode提交结果
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@@ -52,16 +53,17 @@
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```
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- ## Leetcode problem 70 (爬楼梯)
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- 算法思路：动态规划。由于当前状态只取决于前两个状态，所以dp table可以精简为只用两个数表示
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- 算法思路：动态规划。反向遍历，由于当前状态只取决于前两个状态，所以dp table可以精简为只用两个数表示
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- Leetcode提交结果
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- 代码：
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```python
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def climbStairs(self, n: int) -> int:
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pre, cur = 0, 1
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for i in range(1, n + 1):
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pre, cur = cur, pre + cur
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return cur
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class Solution(object):
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def minCostClimbingStairs(self, cost):
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f1 = f2 = 0 # 顶部cost为0
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for x in reversed(cost):
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f1, f2 = x + min(f1, f2), f1
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return min(f1, f2)
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```
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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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- 令`wArr`为重量数组，`vArr`为价值数组，b为背包的最大容量限制，定义`f[i]`为最大容量限制为i的情况下的最大价值，则状态转移方程如下：
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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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$f[b] = max\{f[b], f[b - wArr[i]] + vArr[i]\}$
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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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- 使用一维的dp数组保存中间的计算值
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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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def getMaxValue(wArr, vArr, b):
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# 初始化dp数组
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dp = [0 for _ in range(b + 1)]
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for i in range(len(wArr)):
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for c in range(b, wArr[i] - 1, -1):
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dp[c] = max(dp[c], dp[c - wArr[i]] + vArr[i])
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return dp[b]
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if __name__ == '__main__':
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print(getMaxValue([1, 2, 5, 6, 7], [1, 6, 18, 22, 28], 11))
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# 输出
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40
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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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- ### 投资问题
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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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- 采用动态规划的思想
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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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- 定义$F_k(x)$表示前k个项目投资 x 元的最大收益， $f_k(x)$ 表示第k个项目投资x元的收益
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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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- 则状态转换方程如下：
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- **Leetcode提交结果：**
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$F_k(x) = \begin{cases}\max_{0 \le x_k \le x} \{f_k(x_k) + F_{k - 1}(x - x_k)\}&k > 1\\f_1(x) &k = 1 \end{cases}$
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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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def getMaxProfit(f, x):
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"""
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dp[i][j] 表示前 i + 1 个项目投资j元的最大收益
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:param f:
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:param x:
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:return:
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"""
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columnNum = len(f[0])
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rowNum = len(f)
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dp = [[0 for _ in range(columnNum)] for _ in range(rowNum)]
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dp[0] = f[0]
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for i in range(1, len(f)):
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for j in range(0, x + 1):
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for k in range(0, j + 1):
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if dp[i][j] < f[i][k] + dp[i - 1][j - k]:
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dp[i][j] = f[i][k] + dp[i - 1][j - k]
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return dp[rowNum - 1][columnNum - 1]
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if __name__ == '__main__':
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print(getMaxProfit([
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[0, 11, 12, 13, 14, 15],
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[0, 0, 5, 10, 15, 20],
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[0, 2, 10, 30, 32, 40],
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[0, 20, 21, 22, 23, 24]
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], 5))
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# 输出
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61
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```
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- ### Leetcode problem 120
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- ### Leetcode problem 746
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> [三角形最小路径和](https://leetcode-cn.com/problems/triangle/)
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> [使用最小花费爬楼梯](https://leetcode-cn.com/problems/min-cost-climbing-stairs/)
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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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动态规划。反向遍历，由于当前状态只取决于前两个状态，所以dp table可以精简为只用两个数表示
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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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class Solution(object):
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def minCostClimbingStairs(self, cost):
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f1 = f2 = 0 # 顶部cost为0
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for x in reversed(cost):
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f1, f2 = x + min(f1, f2), f1
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return min(f1, f2)
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```
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