Minimum Path Sum (Java)

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.

经典动态规划题，dp[i][j]数组记录grid从左上角走到i,j位置的最小值，dp[i][j] = grid[i][j] + min{dp[i - 1][j], dp[i][j - 1]}。式子中用到了dp[i - 1][j]和dp[i][j-1]，由于for循环是从1开始的，所以要对dp数组的上边界和左边界赋值。

如果不想特别处理边界就将dp元素初始化为Integer.MAX_VALUE，然后将随便dp[1][0]或dp[0][1]置为0，转移方程改为dp[i][j] = grid[i - 1][j - 1] + min{dp[i - 1][j], dp[i][j-1]}即可。

还有更简单的方法能达到O（n），再写的时候注意下。

Source

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

Test

    public static void main(String[] args){    int[][] grid = {{1,3,1},{1,5,1},{4,2,1}};    System.out.println(new Solution().minPathSum(grid));    }