代码随想录leetcode200题之动态规划算法

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1 介绍

本博客用来记录代码随想录leetcode200题之动态规划算法相关题目。

2 训练

题目1：509. 斐波那契数

C++代码如下，

class Solution {
public:int fib(int n) {if (n <= 1) { //特判return n;}int a = 0, b = 1;for (int i = 2; i <= n; ++i) {int t = a + b;a = b;b = t;}return b;}
};

python3代码如下，

class Solution:def fib(self, n: int) -> int:if n <= 1: #特判return n a, b = 0, 1for i in range(2,n+1):a, b = b, a + breturn b

题目2：70. 爬楼梯

C++代码如下，

class Solution {
public:int climbStairs(int n) {if (n <= 2) { //特判return n;}int a = 1, b = 2;for (int i = 3; i <= n; ++i) {int t = a + b;a = b;b = t;}return b;}
};

python3代码如下，

class Solution:def climbStairs(self, n: int) -> int:if n <= 2: #特判return na, b = 1, 2for i in range(3,n+1):a, b = b, a + breturn b

题目3：746. 使用最小花费爬楼梯

C++代码如下，

class Solution {
public:int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();vector<int> dp(n+1, 0);//状态定义//dp[i]:跑到第i层阶梯所需的最小花费//状态转移//dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])//状态初始化//dp[0] = 0, dp[1] = 0for (int i = 2; i < n+1; ++i) {dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]);}return dp[n];}
};

python3代码如下，

class Solution:def minCostClimbingStairs(self, cost: List[int]) -> int:n = len(cost)dp = [0] * (n+1)#状态定义#dp[i]:爬到第i个阶梯的最小花费#状态转移#dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])#状态初始化#dp[0] = 0, dp[1] = 0for i in range(2, n+1):dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2])return dp[n]

题目4：62. 不同路径

C++代码如下，

数学解法如下，

class Solution {
public:int uniquePaths(int m, int n) {m -= 1;n -= 1;//计算C[n+m][n]long long ans = 1;for (int i = m+1, j = 0; j < n; ++j, ++i) {ans = ans * i / (j + 1);}return ans;}
};

动态规划解法如下，

class Solution {
public:int uniquePaths(int m, int n) {vector<vector<int>> dp(n, vector<int>(m, 0));//状态定义//dp[i][j]:从(0,0)走到(i,j)的走法//状态转移//dp[i][j] = dp[i-1][j] + dp[i][j-1]//状态初始化//dp[i][0] = 1//dp[0][j] = 1for (int i = 0; i < n; ++i) dp[i][0] = 1;for (int j = 0; j < m; ++j) dp[0][j] = 1;for (int i = 1; i < n; ++i) {for (int j = 1; j < m; ++j) {dp[i][j] = dp[i-1][j] + dp[i][j-1]; }}return dp[n-1][m-1];}
};

python3代码如下，

数学解法，

class Solution:def uniquePaths(self, m: int, n: int) -> int:m -= 1n -= 1#计算组合数C[n+m][m]i = n+1j = 1res = 1while j <= m:res = res * i / jj += 1i += 1return int(res)

#C++中可能会超过整型数值范围
class Solution:def uniquePaths(self, m: int, n: int) -> int:m -= 1n -= 1C = [[0] * 110 for _ in range(110)]for i in range(110):for j in range(i+1):if j == 0:C[i][j] = 1else:C[i][j] = C[i-1][j] + C[i-1][j-1]return C[m+n][m]

#C++中可能会超过整型数值范围
class Solution:def uniquePaths(self, m: int, n: int) -> int:m -= 1n -= 1#求C[n+m][m]fa = 1fb = 1for i in range(1,n+1):fa *= m + n - i + 1fb *= i res = fa // fbreturn res

动态规划解法

class Solution:def uniquePaths(self, m: int, n: int) -> int:dp = [[0] * m for _ in range(n)]dp[0][0] = 1for i in range(1,n):dp[i][0] = dp[i-1][0]for j in range(1,m):dp[0][j] = dp[0][j-1]for i in range(1,n):for j in range(1,m):dp[i][j] = dp[i-1][j] + dp[i][j-1]return dp[n-1][m-1]

题目5：63. 不同路径 II

C++代码如下，

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int n = obstacleGrid.size();int m = obstacleGrid[0].size();vector<vector<int>> dp(n, vector<int>(m, 0));dp[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;for (int i = 1; i < n; ++i) {if (obstacleGrid[i][0] == 0) {dp[i][0] = dp[i-1][0];}}for (int j = 1; j < m; ++j) {if (obstacleGrid[0][j] == 0) {dp[0][j] = dp[0][j-1];}}for (int i = 1; i < n; ++i) {for (int j = 1; j < m; ++j) {if (obstacleGrid[i][j] == 0) {dp[i][j] = dp[i-1][j] + dp[i][j-1];}}}return dp[n-1][m-1];}
};

python3代码如下，

class Solution:def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:n = len(obstacleGrid)m = len(obstacleGrid[0])dp = [[0] * m for _ in range(n)]dp[0][0] = 1 if obstacleGrid[0][0] == 0 else 0for i in range(1,n):if obstacleGrid[i][0] == 0:dp[i][0] = dp[i-1][0]for j in range(1,m):if obstacleGrid[0][j] == 0:dp[0][j] = dp[0][j-1]for i in range(1,n):for j in range(1,m):if obstacleGrid[i][j] == 0:dp[i][j] = dp[i-1][j] + dp[i][j-1]return dp[n-1][m-1]

题目6：343. 整数拆分

C++代码如下，

class Solution {
public:int integerBreak(int n) {int res = 1;for (int k = 2; k <= n; ++k) {//将n拆分成k个数int a = n / k;vector<int> nums(k, a);int t = n - a * k;for (int i = 0; i < t; ++i) nums[i] += 1;int ans = 1;for (auto x : nums) ans *= x;res = max(res, ans);}return res;}
};

python3代码如下，

class Solution:def integerBreak(self, n: int) -> int:res = 1 #答案的理论最小值for k in range(2,n):#将n拆分成k个数a = n // knums = [a] * kt = n - a * kfor i in range(t):nums[i] += 1ans = 1for x in nums:ans *= x res = max(res, ans)return res

题目7：96. 不同的二叉搜索树

C++代码如下，

class Solution {
public:int numTrees(int n) {vector<int> dp(n + 1, 0);dp[0] = 1;for (int i = 1; i <= n; ++i) {for (int j = 1; j <= i; ++j) {dp[i] += dp[j-1] * dp[i-j];}}return dp[n];}
};

python3代码如下，

class Solution:def numTrees(self, n: int) -> int:dp = [0] * (n + 1) dp[0] = 1for i in range(n+1):for j in range(1,i+1):dp[i] += dp[j-1] * dp[i-j]return dp[n]

题目8：46. 携带研究材料（第六期模拟笔试）

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 5010, M = 5010;
int v[N];
int w[N];
int m;
int n;
int dp[M];int main() {cin >> n >> m;for (int i = 0; i < n; ++i) {cin >> v[i];}for (int i = 0; i < n; ++i) {cin >> w[i];}for (int i = 0; i < n; ++i) {for (int j = M; j - v[i] >= 0; --j) {dp[j] = max(dp[j], dp[j-v[i]] + w[i]);}}cout << dp[m] << endl;return 0;
}

python3代码如下，

import sys data = sys.stdin.read()
lines = data.splitlines()
n, m = map(int, lines[0].split())
v = list(map(int, lines[1].split()))
w = list(map(int, lines[2].split()))dp = [0] * (m + 10)
for i in range(n):for j in range(m,v[i]-1,-1):dp[j] = max(dp[j], dp[j-v[i]] + w[i])
print(dp[m])

题目9：416. 分割等和子集

C++代码如下，

class Solution {
public:bool canPartition(vector<int>& nums) {int totalSum = accumulate(nums.begin(), nums.end(), 0);if (totalSum % 2 != 0) return false;int targetSum = totalSum / 2;vector<int> dp(targetSum + 10, 0);for (int i = 0; i < nums.size(); ++i) {for (int j = targetSum; j >= nums[i]; --j) {dp[j] = max(dp[j], dp[j-nums[i]] + nums[i]);}}return dp[targetSum] == targetSum;}
};

python3代码如下，

class Solution:def canPartition(self, nums: List[int]) -> bool:totalSum = sum(nums)if totalSum % 2 != 0:return FalsetargetSum = totalSum // 2#从nums中能否找到一些数，使得它们之和等于targetSum#套用01背包模型dp = [0] * (targetSum + 10)for i in range(len(nums)):for j in range(targetSum, nums[i]-1, -1):dp[j] = max(dp[j], dp[j-nums[i]] + nums[i])return dp[targetSum] == targetSum