【LeetCode】Triangle

最新推荐文章于 2020-05-23 10:06:18 发布

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本文介绍了一种使用动态规划解决三角形最小路径和问题的方法。通过递推公式dp[i][j]=min(dp[i-1][j-1], dp[i-1][j])+val(i,j)，实现了从顶到底的最小路径求和，特别处理了每行的第一个和最后一个元素。

题目描述：

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.

解：

用动态规划，dp[i][j]表示到达（i，j）点的最小和，则有dp[i][j]=min(dp[i-1][j-1]，dp[i-1][j])+val(i，j) ，行首和行尾要拎出来单独处理。

代码如下：

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