leetcode--Minimum Path Sum

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

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

题意：给定m*n矩阵，里面都是非负数，求从左上角，走到右下角的路径，要求这条路径过程中所有数加起来的和最小

分类：数组，动态规划

解法1：显然是动态规划，保留上次的状态，对于每一个位置(i,j)

从左上角到这个位置的最小和，由(i-1,j)和(i,j-1)决定

[java] view
plain copy

public class Solution {  

    public int minPathSum(int[][] grid) {  

        int m = grid.length;  

        int n = grid[0].length;  

        int[][] flag = new int[m]
;  

        for(int i=0;i<m;i++){  

            for(int j=0;j<n;j++){  

                if(i==0&&j==0) flag[0][0] = grid[0][0];  

                else if(i==0) flag[i][j] = flag[0][j-1]+grid[i][j];  

                else if(j==0) flag[i][j] = flag[i-1][0]+grid[i][j];  

                else flag[i][j] = Math.min(flag[i-1][j],flag[i][j-1])+grid[i][j];  

            }  

        }  

        return flag[m-1][n-1];  

    }  

}  

[java] view
plain copy

public class Solution {  

    public int minPathSum(int[][] grid) {  

        int m = grid.length;  

        int n = grid[0].length;  

        int[][] flag = new int[m]
;  

          

        for(int i=m-1;i>=0;i--){  

            for(int j=n-1;j>=0;j--){               

                if(i==m-1&&j==n-1){  

                    flag[i][j] = grid[m-1][n-1];  

                }else if(i==m-1){  

                    flag[i][j] = flag[i][j+1]+grid[i][j];  

                }else if(j==n-1){  

                    flag[i][j] = flag[i+1][j]+grid[i][j];  

                }else{  

                    flag[i][j] = Math.min(flag[i+1][j],flag[i][j+1])+grid[i][j];  

                }  

            }  

        }  

        return flag[0][0];  

    }  

}  

原文链接http://blog.csdn.net/crazy__chen/article/details/46409705