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.

• 大意就是給定一個m*n的矩陣，只能向下or向右前進，找到到達右下角最小的路徑和。

• 典型的動態(tài)規(guī)劃問題：
  注意：貪心策略是不可行的，在本問題中，局部的最優(yōu)解無法得到最優(yōu)解，因此需要使用dp，綜合并記錄全局的所有局部最優(yōu)路徑，最終得到最優(yōu)解。

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

    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數(shù)組的第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];

        //狀態(tài)轉(zhuǎn)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];
    }