[LeetCode] 756. Pyramid Transition Matrix

题目描述

We are stacking blocks to form a pyramid. Each block has a color which is a one letter string, like 'Z'.

For every block of color C we place not in the bottom row, we are placing it on top of a left block of color A and right block of color B. We are allowed to place the block there only if (A, B, C) is an allowed triple.

We start with a bottom row of bottom, represented as a single string. We also start with a list of allowed triples allowed. Each allowed triple is represented as a string of length 3.

Return true if we can build the pyramid all the way to the top, otherwise false.

Example 1:

Input: bottom = "XYZ", allowed = ["XYD", "YZE", "DEA", "FFF"]
Output: true
Explanation:
We can stack the pyramid like this:
    A
   / \
  D   E
 / \ / \
X   Y   Z

This works because ('X', 'Y', 'D'), ('Y', 'Z', 'E'), and ('D', 'E', 'A') are allowed triples.

Example 2:

Input: bottom = "XXYX", allowed = ["XXX", "XXY", "XYX", "XYY", "YXZ"]
Output: false
Explanation:
We can't stack the pyramid to the top.
Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D.

Note:
bottom will be a string with length in range [2, 8].
allowed will have length in range [0, 200].
Letters in all strings will be chosen from the set {‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’}.

输入是一个叫bottom 的字符串以及allowed 数组，bottom 字符串代表金字塔的最底层，allowed 数组给出了所有可能的金字塔的构建方法。给定这两个数组后判断是否能建成金字塔。
解题思路是使用递归的方式，每次递归构建出上一层，递归的终止条件是当前层只有一个字符。
需要注意的是，每次递归时需要使用dfs 搜索找出当前层所有可能的构造方式。具体见实现以及注释。

C++ 实现

class Solution {
public:
    bool pyramidTransition(string bottom, vector<string>& allowed) {
        // 构建allowed_map, hash_map 的key是小金子塔的底部，value 是所有允许的顶部的集合
        for (const string& str : allowed)
        {
            const string& key = str.substr(0, 2);
            const char value = str.back();
            if (allowed_map.count(key))
            {
                allowed_map[key].insert(value);
            }
            else
            {
                allowed_map[key] = {value};
            }
        }
        // 递归搜索
        return dfs(bottom);
    }
private:
    unordered_map<string, set<char>> allowed_map;

    bool dfs(const string& bottom)
    {
        // 终止条件
        if (bottom.size() == 1) return true;
        // 首先判断上一层无法构建的情况，如果无法构建直接return false
        for(size_t i = 0; i < bottom.size() - 1; i++)
        {
            if (!allowed_map.count(bottom.substr(i, 2)))
            {
                return false;
            }
        }
        vector<string> candidates;
        candidates.reserve(bottom.size() - 1);
        string path;
        //使用dfs的方式找到上一层所有可能结果
        getCandidates(bottom, 0, path, candidates);
        //遍历所有结果进行递归搜索
        for(const string& candidate : candidates)
        {
            if (dfs(candidate))
            {
                return true;
            }
        }
        return false;
    }
    // dfs的方式找到所有可能的上一层的结果
    // 结果存储到 candidates 数组中
    // cur 存储当前dfs的进度
    void getCandidates(const string& bottom, 
                       size_t idx,
                       string& cur,
                       vector<string>& candidates)
    {
        if (idx == bottom.size() - 1)
        {
            candidates.push_back(cur);
            return;
        }

        for(const char c : allowed_map.at(bottom.substr(idx, 2)))
        {
            cur += c;
            getCandidates(bottom, idx + 1, cur, candidates);
            cur.pop_back();
        }
    }
};