Minimum Path Sum leetcode

最小路径和：二维与一维动态规划解法

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签：数据结构与算法

原文链接：https://segmentfault.com/a/1190000004296866

本文详细介绍了如何使用二维和一维动态规划解决最小路径和问题，包括状态转移方程、初始化步骤及代码实现，旨在帮助读者理解动态规划在解决实际问题中的应用。

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.

二维动态规划

说明

state: dp[x][y]从起点走到x,y的最短路径
function: dp[x][y] = min(dp[x-1][y], dp[x][y-1]) + cost[x][y]
intialize: dp[0][0] = cost[0][0]
// f[i][0] = sum(0,0 -> i,0)
// f[0][i] = sum(0,0 -> 0,i)
answer: f[n-1][m-1]

复杂度

时间O(mn) 空间O(mn)

代码

public int minPathSum(int[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    int[][] dp = new int[m][n];
    dp[0][0] = grid[0][0];
    for (int i = 1; i < n; i++) {
        dp[0][i] = grid[0][i] + dp[0][i - 1];
    }
    for (int j = 1; j < m; j++) {
        dp[j][0] = grid[j][0] + dp[j - 1][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];
}

一维动态规划

说明

初始化第一行的数据, 然后每次遍历一行覆盖一次数据, 思路与二维相同

复杂度

时间O(m*n) 空间O(n)

代码

public int minPathSum(int[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    int[] dp = new int[n];
    dp[0] = grid[0][0];
    for (int i = 1; i < n; i++) {
        dp[i] = dp[i - 1] + grid[0][i];
    }
    for (int i = 1; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (j != 0) {
                dp[j] = Math.min(dp[j], dp[j - 1]);
            }
            dp[j] += grid[i][j];
        }
    }
    return dp[n - 1];
}