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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