[LeetCode]Minimum Path Sum,解题报告

前言

这道题目我今年面试的时候考过，不给出具体的哪家公司了，也是给定矩阵从左上角到右下角的和最小的路径。

开始我并不知道是确定了起始点和结束点，因此我第一反应是用DFS遍历矩阵，然后那个面试官说不让我用递归（其实dfs也不一定非用递归实现）。想了一下我说用动态规划，给他写了状态方程，时间复杂度为O(n^2)，他非跟我纠结用动态规划可以到O(n)的时间复杂度，我表示无语。虽然最后我还是拿到了这家的offer，不过面试官的水平让我有些失望，最终还是没去

题目

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.

思路

很简单的二维动态规划题目，我直接给出状态方程，设dp[i][j]为从(0,0)到(i,j)的路径和最小值

dp[i][j] = MIN(dp[i - 1][j], dp[i][j - 1]) + matrix[i][j]

AC代码

import java.util.Scanner;public class MinimunPathSum {    public static void main(String[] args) {        int i, j, m, n, sum, grid[][];        Scanner cin = new Scanner(System.in);        while (cin.hasNext()) {            m = cin.nextInt();            n = cin.nextInt();            grid = new int[m][n];            for (i = 0; i < m; i++) {                for (j = 0; j < n; j++) {                    grid[i][j] = cin.nextInt();                }            }            sum = minPathSum(grid);            System.out.println(sum);        }        cin.close();    }    public static int minPathSum(int[][] grid) {        int i, j, dp[][] = new int[grid.length][grid[0].length];        int col, row;        // initial variable        row = grid.length;        col = grid[0].length;        // initial dp array        dp[0][0] = grid[0][0];        for (i = 1; i < row; i++) {            dp[i][0] = dp[i - 1][0] + grid[i][0];        }        for (j = 1; j < col; j++) {            dp[0][j] = dp[0][j - 1] + grid[0][j];        }        // dynamic process, dp[i][j] = MIN{dp[i - 1][j], dp[i][j - 1]} + grid[i][j]        for (i = 1; i < row; i++) {            for (j = 1; j < col; j++) {                dp[i][j] =                        dp[i - 1][j] <= dp[i][j - 1] ? dp[i - 1][j] + grid[i][j] : dp[i][j - 1]                                + grid[i][j];            }        }        return dp[row - 1][col - 1];    }}