[LeetCode]Triangle

最新推荐文章于 2020-05-23 10:06:18 发布

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本文介绍了一个使用动态规划解决三角形网格中寻找最小路径和的问题。通过详细的C++代码实现展示了如何初始化状态转移方程，并逐步迭代求解最终的最小路径总和。

#define INF 0x6FFFFFFF
class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		if(triangle.size() == 0)
			return 0;
		int n = triangle.size();
        vector<vector<int>> dp(n+1);
		for(int i = 0; i <= n; ++i)
			dp[i].assign(n+1, INF);
		//init value
		dp[1][1] = triangle[0][0];
		for(int i = 2; i <= n; ++i)
		{
			for(int j = 1; j <= i; ++j)
			{
				dp[i][j] = min(dp[i-1][j-1], dp[i-1][j])+triangle[i-1][j-1];
			}
		}
		//output
		int ans = INF;
		for(int j = 1; j <= n; ++j)
			ans = min(dp[n][j], ans);
		return ans;
    }
};

second time

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = triangle.size();
        if(n == 0) return 0;
        vector<int> f(n+1, INT_MAX);
        for(int m = 1; m <= n; ++m)
        {
            for(int j = m; j >= 1; --j)
            {
                if(m == 1 && j == 1) f[j] = triangle[m-1][j-1];
                else f[j] = min(f[j], f[j-1])+triangle[m-1][j-1];
            }
        }
        
        int mmin = INT_MAX;
        for(int j = 1; j <= n; ++j) mmin = min(f[j], mmin);
        
        return mmin;
    }
};