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.

• 大意就是给定一个m*n的矩阵，只能向下or向右前进，找到到达右下角最小的路径和。

• 典型的动态规划问题：
  注意：贪心策略是不可行的，在本问题中，局部的最优解无法得到最优解，因此需要使用dp，综合并记录全局的所有局部最优路径，最终得到最优解。

• 递推公式：dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]

演示 :	grid[][]
1	2
3	4
5	6

演示 :	dp[][]
1	3
4	7
9	13

    public static int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];

        //初始化dp[0][0]
        dp[0][0] = grid[0][0];
        //初始化dp数组的第0行和0列
        for (int i = 1; i < m; ++i)
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        for (int i = 1; i < n; ++i)
            dp[0][i] = dp[0][i - 1] + grid[0][i];

        //状态转移公式：dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
        for (int i = 1; i < m; ++i)
            for (int j = 1; j < n; ++j)
                    dp[i][j]= Math.min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j];

        return dp[m - 1][n - 1];
    }