DP路径的数目的系列问题

class Solution {
         public int uniquePaths(int m, int n) {
        if(m==1&&n==1){
            return 1;
        }
        int dp[][]=new int[m][n];
        for (int i = 0; i < n; i++) {
            dp[0][i]=1;
        }
        for (int i = 1; i < m; i++) {
            dp[i][0]=1;
            for (int j = 1; j < n; j++) {
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }



}

暴力枚举

 public int minPathSum(int[][] grid) {
        int n=grid.length;//行
        int m=grid[0].length;//列
        if (n==0&&m==0){
            return 0;
        }
        return helper(n,m,grid);
    }

    
    private int helper(int n, int m, int[][] grid) {
        if(n==0&&m==0){
            return grid[0][0];
        }else if(n==0){
            return helper(n,m-1,grid)+grid[n][m];
        }else if(m==0){
            return helper(n-1,m,grid)+grid[n][m];
        }
        return Math.min(helper(n-1,m,grid),helper(n,m-1,grid))+grid[n][m];

    }

 优化

剪枝

class Solution {
 public int minPathSum(int[][] grid) {
        int n=grid.length;//行
        int m=grid[0].length;//列
        int dp[][]=new int[n+1][m+1];
        for (int rows[]: dp) {
            Arrays.fill(rows,-1);
        }
        if (n==0&&m==0){
            return 0;
        }
        return helper(n-1,m-1,grid,dp);
    }

    
    private int helper(int n, int m, int[][] grid, int[][] dp) {
        if(dp[n][m]!=-1){
            return dp[n][m];
        }
        if(n==0&&m==0){
            return grid[0][0];
        }else if(n==0){
            dp[n][m]=helper(n,m-1,grid, dp)+grid[n][m];
            return dp[n][m];
        }else if(m==0){
            dp[n][m]=helper(n-1,m,grid, dp)+grid[n][m];
            return dp[n][m];
        }
        dp[n][m]=Math.min(helper(n-1,m,grid, dp),helper(n,m-1,grid, dp))+grid[n][m];
        return dp[n][m];

    }
}

自底向上的动态规划

class Solution {
         public int minPathSum(int[][] grid) {
        //base case 
        int n=grid.length;
        int m=grid[0].length;
        int dp[][]=new int[n+1][m+1];//左上角到[i][j]的最短步数（带权值）
        //base case
        dp[0][0]=grid[0][0];
        for (int i = 1; i < m; i++) {
            dp[0][i]=dp[0][i-1]+grid[0][i];
        }
        for (int i = 1; i < n; i++) {
            dp[i][0]=dp[i-1][0]+grid[i][0];
        }
        for (int i = 1; i