动态规划Day01

TheLegendMe2025-12-02 11:24

cpp 复制代码

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

        for(int i = 0; i < m && obstacleGrid[i][0] == 0; i ++ ) dp[i][0] = 1;
        for(int j = 0; j < n && obstacleGrid[0][j] == 0; j ++ ) dp[0][j] = 1;

        for(int i = 1 ; i < m ; i ++ ) {
            for(int j = 1; j < n; j ++ ) {
                if(obstacleGrid[i][j] == 1) continue;
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m -1][n -1];
    }
};

五步走：1确定dp[i]的意义2找递推公式3确定初始值4确定遍历方向5代值验算

动态规划：343.整数拆分

cpp 复制代码

class Solution {
public:
    int integerBreak(int n) {
        vector<int>dp(n+1);
        dp[2] = 1;
        for(int i = 3; i <= n; i ++ ) {
            for(int j = 1; j <= i/2; j ++ ) {
                dp[i] = max(dp[i],max(j * (i - j),j * dp[i - j]));
            }
        }
        return dp[n];
    }
};

3，如何确定大小端：

cpp 复制代码

#include <iostream>
#include <cstring>

using namespace std;

bool test() {
    union {
        uint32_t i;
        uint8_t c[4];
    }u;
    u.i = 1;
    return u.c[0] == 1;
}

int main() {
    if(test()) {
        cout<<"small"<<endl;
    } else {
        cout <<"big" <<endl;
    }
    return 0;
}