动态规划——minimum-path-sum

题目描述

一个 m x n 的矩阵填充着非负整数，找到从左上角到右下角和最小的路径。

注：每一步只能向下一格走或向右一格走。

递归例程：

public class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null||grid.length == 0||grid[0].length == 0)
            return 0;
        return help(grid,0,0);
    }
    public int help(int[][] grid,int row,int col)
        {
        if(row == grid.length-1&&col == grid[0].length-1)
            return grid[row][col];//达到bottom right时返回该值
        
        if(row == grid.length-1&&col!=grid[0].length-1)//到达最后一行的情况
            return grid[row][col]+help(grid,row,col+1);
        if(col == grid[0].length-1&&row!=grid.length-1)//到达最后一列的情况
            return grid[row][col]+help(grid,row+1,col);
        else//一般情况
            {
        int down=help(grid,row+1,col);
        int right=help(grid,row,col+1);
        if(down<right)
            return down+grid[row][col];
        else
            return right+grid[row][col];
        }
    }
}

DP例程：

public class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null||grid.length == 0||grid[0].length == 0)
            return 0;
        int row=grid.length;
        int col=grid[0].length;
       int[][] dp=new int[row][col];
        //第一个值初始化
        dp[row-1][col-1]=grid[row-1][col-1];
        //对最下行初始化 
        for(int j=col-2;j>=0;j--)
            {
            dp[row-1][j]=dp[row-1][j+1]+grid[row-1][j];
        }
        //对最右列初始化
        for(int i=row-2;i>=0;i--)
            {
            dp[i][col-1]=dp[i+1][col-1]+grid[i][col-1];
        }
        //general
        for(int i=row-2;i>=0;i--)
            {
            for(int j=col-2;j>=0;j--)
                {
                int down=dp[i+1][j];
                int right=dp[i][j+1];
                dp[i][j]=down<right?down+grid[i][j]:right+grid[i][j];
            }
        }
        return dp[0][0];
    }
}

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