Minimum Path Sum

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

Solution:

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

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