【c++】【leetcode62】不同路径1/2,路径和(动态规划)

不同路径

使用动态规划算法

• dp[i][j]代表走到第i行第j列有多少路径
• dp[0][j]只能由dp[0][j-1]走来dp[i][0]只能由dp[i-1][0]走来

if(i == 0 && j != 0)dp[i][j] = dp[i][j - 1];
else if(j == 0 && i != 0)dp[i][j] = dp[i - 1][j];
else if(!i && !j)continue;
else dp[i][j] = dp[i-1][j] + dp[i][j-1];

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector> dp(m,vector(n));
        dp[0][0] = 1;
        //初始化dp[0][j]只能由dp[0][j-1]走来dp[i][0]只能由dp[i-1][0]走来
        for(int i = 0;i < m;++i){
            for(int j = 0;j < n;++j){
                if(i == 0 && j != 0)dp[i][j] = dp[i][j - 1];
                else if(j == 0 && i != 0)dp[i][j] = dp[i - 1][j];
                else if(!i && !j)continue;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];       
    }
};
//dp[i][j]代表走到第i行第j列有多少路径

不同路径2

对比1添加了不同的条件：由障碍物

• 遇到障碍物直接标为0就可以
• 如果开始和结尾有一个是障碍物，那就无法走到，直接返回0

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

        for(int i = 0;i < m;++i){
            for(int j = 0;j < n;++j){
                if(i == 0 && j == 0)continue;
                if(obstacleGrid[i][j])dp[i][j] = 0;
                else if(i == 0)dp[i][j] = dp[i][j - 1];
                else if(j == 0)dp[i][j] = dp[i-1][j];
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};

最小路径和

同上

class Solution {
public:
    int minPathSum(vector>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector> dp(m,vector(n));
        dp[0][0] = grid[0][0];

        for(int i = 0;i < m;++i){
            for(int j = 0;j < n;++j){
                if(i == 0 && j == 0)continue;
                if(i == 0)dp[i][j] = dp[i][j-1] + grid[i][j];
                else if(j == 0)dp[i][j] = dp[i - 1][j] + grid[i][j];
                else dp[i][j] = min(dp[i][j-1],dp[i - 1][j]) + grid[i][j];
            }
        } 
        return dp[m-1][n-1];
    }
};