【DP】LeetCode 120. Triangle

最新推荐文章于 2024-06-27 13:46:32 发布

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#LeetCode 120. Triangle

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本文提供了一种解决LeetCode 120题“Triangle”最小路径和问题的有效算法实现。通过两次迭代改进，最终的代码简化了原始思路，实现了对输入三角形数组的动态规划处理，有效地找到了从顶点到底部的最小路径总和。

LeetCode 120. Triangle

Solution1：我的答案
真费劲啊！！！
啊！！！！！！

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if (!triangle.size() 
            || !triangle[0].size()) // || 是短路运算符 
            return 0;
        int m = triangle.size(), min_path = INT_MAX;
        for (int i = 0; i < m - 1; i++) {
            for (int j = 0; j < triangle[i].size(); j++) {
                if (i == 0) {
                    triangle[1][0] += triangle[0][0];
                    triangle[1][1] += triangle[0][0];
                }
                else if (j == 0)
                    triangle[i+1][j] += triangle[i][j];
                else if (j == triangle[i].size() - 1) {
                    triangle[i+1][j+1] += triangle[i][j];
                    triangle[i+1][j] += min(triangle[i][j-1], 
                                            triangle[i][j]);
                }
                else {
                    triangle[i+1][j] += min(triangle[i][j-1], 
                                            triangle[i][j]);
                }
            }
        }
        for (int i = 0; i < triangle[m-1].size(); i++)
            min_path = min(min_path, triangle[m-1][i]);
        return min_path;
    }
};

Solution2：我的答案改进版
这个代码只对等边三角形有效，但是也能AC了~~~

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if (!triangle.size() 
            || !triangle[0].size()) // || 是短路运算符 
            return 0;
        int m = triangle.size(), min_path = INT_MAX;
        for (int i = 1; i < m; i++) {
            for (int j = 0; j < triangle[i].size(); j++) {
                if (j == 0)
                    triangle[i][j] += triangle[i-1][j];
                else if (j == triangle[i].size() - 1) 
                    triangle[i][j] += triangle[i-1][j-1];
                else
                    triangle[i][j] += min(triangle[i-1][j-1], 
                                            triangle[i-1][j]);
            }
        }
        for (int i = 0; i < triangle[m-1].size(); i++)
            min_path = min(min_path, triangle[m-1][i]);
        return min_path;
    }
};