64. Minimum Path Sum

最新推荐文章于 2025-08-11 09:20:14 发布

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最新推荐文章于 2025-08-11 09:20:14 发布

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文章标签： java 数据结构与算法 runtime

原文链接：https://segmentfault.com/a/1190000018125245

本文详细解析了在给定的m*n网格中寻找从左上角到右下角的最小路径和问题。通过动态规划的方法，阐述了如何利用状态转移方程grid[i][j]=grid[i][j]+min(grid[i-1][j],grid[i][j-1])来高效求解此问题，最终实现了4ms的运行时间，优于97.83%的Java在线提交。

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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.

难度：medium

题目：给定m * n的格子，找出从左上角到右下角和最小的路径。

思路：动态规划，状态转移方程
grid[i][j] = grid[i][j] + min(grid[i - 1][j], grid[i][j - 1]

Runtime: 4 ms, faster than 97.83% of Java online submissions for Minimum Path Sum.
Memory Usage: 40.8 MB, less than 1.47% of Java online submissions for Minimum Path Sum.

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