[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]
;

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];
}
}