LeetCode: Triangle

最新推荐文章于 2019-07-27 09:27:00 发布

原创最新推荐文章于 2019-07-27 09:27:00 发布 · 384 阅读

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本文探讨了如何通过动态规划算法解决三角形中从顶点到底边的最小路径和问题，通过实例展示了求解过程及关键步骤。

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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.

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        int sizeX = triangle.size();
        if(sizeX == 0)
            return 0;
        int sizeY = triangle[sizeX-1].size();
        vector<int> cur(sizeY);
        for(int x = sizeX - 1; x >=0; x--)
        {
            for(int y = 0; y < triangle[x].size(); y++)
            {
                if(x == sizeX - 1)
                {
                    cur[y] = triangle[x][y];
                }
                else
                {
                    cur[y] = std::min(cur[y], cur[y+1]) + triangle[x][y];   
                }
            }
        }
        return cur[0];
        
    }
};

Round 2:

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