[LeetCode] 120. Triangle

最新推荐文章于 2024-06-27 13:46:32 发布

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最新推荐文章于 2024-06-27 13:46:32 发布

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原文链接：https://segmentfault.com/a/1190000017160459

本文探讨了如何在给定的三角形结构中找到从顶到底部的最小路径总和，通过动态规划方法，实现了一个高效的解决方案，仅使用O(n)额外空间。

Problem

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.

Bottom-top DP

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

Non Extra Space DP

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int len = triangle.size();
        for (int i = len-2; i >= 0; i--) {
            for (int j = 0; j <= i; j++) {
                int preMin = Math.min(triangle.get(i+1).get(j), triangle.get(i+1).get(j+1));
                int curMin = preMin + triangle.get(i).get(j);
                triangle.get(i).set(j, curMin);
            }
        }
        return triangle.get(0).get(0);
    }
}