120. Triangle

最新推荐文章于 2025-08-22 13:45:39 发布

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最新推荐文章于 2025-08-22 13:45:39 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/1190000018177568

本文探讨了如何在给定的三角形数组中找到从顶部到底部的最小路径总和，详细介绍了使用动态规划解决这一问题的方法，并提供了一个Java实现案例，该算法在在线提交中表现出色。

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Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

难度：medium

题目：给定一三角形数组，找出从上到下最小的路径和。每步只可以向下一行的相邻元素移动。

思路：动态规划

Runtime: 4 ms, faster than 87.60% of Java online submissions for Triangle.
Memory Usage: 38.5 MB, less than 100.00% of Java online submissions for Triangle.

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int m = triangle.size();
        if (1 == m) {
            return triangle.get(0).get(0);
        }
        
        int[][] table = new int[m][m];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j <= i; j++) {
                table[i][j] = triangle.get(i).get(j);
            }
        }
        int result = table[0][0];
        for (int i = 1; i < m; i++) {
            result = table[i][0] + table[i - 1][0];
            for (int j = 0; j <= i; j++) {
                if (0 == j) {
                    table[i][j] += table[i - 1][j];
                } else if (j == i) {
                    table[i][j] += table[i - 1][j - 1];
                } else {
                    table[i][j] += Math.min(table[i - 1][j], table[i - 1][j - 1]);
                }
                result = Math.min(table[i][j], result);
            }
        }
        return result;
    }
}