题目描述

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.

算法

(动态规划) O(m*n)

实在比较简单，不写过程了, 思路就是转移时候每个点[i,j] 只会从 [i-1, j] 和 [i, j-1]转移过来。
定义二维数组dp[i][j]就可以了。
空间复杂度先不优化了。

Java 代码

public class Solution {

    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }

        int m = grid.length;
        int n = grid[0].length;

        // 表示(0, 0)走到(m-1, n-1)的路径最小数字总和
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];

        // 第一行
        for (int j = 1; j < n; j++) {
            dp[0][j] = dp[0][j - 1] + grid[0][j];
        }

        // 第一列
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }

        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];
    }
}