Minimum Path Sum

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Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

public class Solution {    public int minPathSum(int[][] grid) {        if(grid==null) return 0;        return dp(grid);    }    public int dp(int[][] grid){        int m=grid.length;        int n=grid[0].length;        int [][]path=new int[m][n];        path[0][0]=grid[0][0];        for(int i=1;i<m;i++)             path[i][0]=path[i-1][0]+grid[i][0];        for(int j=1;j<n;j++)             path[0][j]=path[0][j-1]+grid[0][j];        for(int i=1;i<m;i++){            for(int j=1;j<n;j++){                path[i][j]=Math.min(path[i-1][j],path[i][j-1])+grid[i][j];            }        }        return path[m-1][n-1];    }}

思路：dp，用递归会超时。

public class Solution {    public int minPathSum(int[][] grid) {        if(grid==null) return 0;        return dp(grid);    }    public int dp(int[][] grid){        int m=grid.length;        int n=grid[0].length;        int []f=new int[n];        f[0]=grid[0][0];        for(int i=1;i<n;i++) f[i]=f[i-1]+grid[0][i];                for(int i=1;i<m;i++){            f[0]+=grid[i][0];            for(int j=1;j<n;j++){                f[j]=Math.min(f[j],f[j-1])+grid[i][j];            }        }        return f[n-1];    }}

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