Leetcode115: Triangle

最新推荐文章于 2025-09-06 17:22:57 发布

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最新推荐文章于 2025-09-06 17:22:57 发布

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分类专栏： leetcode c++ 文章标签： algorithm leetcode

本文链接：https://blog.youkuaiyun.com/u013089961/article/details/49364501

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本文介绍了一种求解三角形最小路径和的问题，通过动态规划的方法从顶点到底部找到一条路径使得路径上的数值之和最小。文章提供了两种解决方案：一种使用额外的空间存储中间结果；另一种直接修改原数组，节省了空间。

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.

要求空间复杂度不超过n

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

另外看到一个更神的解法，不用开辟空间，只是原数组会修改。

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