LeetCode:Triangle

最新推荐文章于 2025-08-10 09:12:36 发布

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本文探讨了如何使用动态规划解决寻找三角形中从顶点到底部最小路径和的问题，并提供了优化后的代码实现。通过滚动数组技术，将空间复杂度优化至O(n)，并提供了解题思路及简洁的代码示例。

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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 onlyO(n) extra space, wheren is the total number of

rows in the triangle.

解题思路:

首先dp是很容易想到的，用dp[i][j]表示到第i层j列的最小值，那么状态转移为:

dp[i][j]=min(dp[i-1][j],dp[i-1][j-1])+triangle[i][j] ( 0 < j < triangle[i].size() - 1).由于状态转移只与i和i-1

行有关，所以用滚动数组能将其内存优化为O(n)(此处有常系数2).关于内存的优化，其实是可以做

到严格的O(n)的，我们通过反向更新即可实现.

解题代码:

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) 
    {
        const int n = triangle.size();
        int dp[n];
        dp[0] = triangle[0][0];
        for (int i = 1; i < n; ++i)
        {
            for (int j = triangle[i].size() - 1; j >= 0 ; --j)
            {
                if (j == triangle[i].size() - 1 || !j)
                    dp[j] = (!j ? dp[0] : dp[j-1]) + triangle[i][j];
                else
                    dp[j] = min(dp[j],dp[j-1]) + triangle[i][j];
            }
        }
        return *min_element(dp,dp+n);
    }
};

更简单的代码:

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) 
    {
        const int n = triangle.size();
        int dp[n];
        for (int i = n -1; i >= 0; --i)
            dp[i] = triangle[n-1][i];
        for (int i = n -2; i >= 0; --i)
            for (int j = 0; j < triangle[i].size(); ++j)
                dp[j] = min(dp[j],dp[j+1]) + triangle[i][j];
        return dp[0];
    }
};