[LeetCode每日一题]64. Minimum Path Sum

题目如下：

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.

64. Minimum Path Sum 给定一个m*n的二维矩阵，让我们找到从左上角到右下角的最小路径和。这道题可以用动态规划解决。设在grid[i][j]的最小路径和为dp[i][j],状态转移方程为：

dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]      i > 0, j > 0

实现代码如下：

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