Pyramid Transition Matrix

问题：

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:

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

解决：

【题意】

有一个类似于杨辉三角的金字塔三角形，给出了底部，还有一些规则。规则的格式是（ABC），C的左儿子允许A出现，C的右儿子允许B出现。根据底部和这些规则，求出能不能构造出一个金字塔三角形

①  使用map记录所有的左右子节点与根节点的组合，dfs遍历查找根节点，记录每一层的字符串，直到最后一层字符串长度为1为止。

class Solution { //11ms
    public boolean pyramidTransition(String bottom, List<String> allowed) {
        Map<String,List<String>> map = new HashMap<>();//键为左右节点字符串，值为根节点的集合
        for (String str : allowed){
            String key = str.substring(0,2);
            if (! map.containsKey(key)){
                map.put(key,new ArrayList<>());
            }
            map.get(key).add(str.substring(2));
        }
        return dfs(map,bottom,new StringBuilder(),0);
    }
    public boolean dfs(Map<String,List<String>> map,String bottom,StringBuilder nextBottom,int p){
        if (p == bottom.length() - 1){//完成一行，更新bottom和p
            bottom = nextBottom.toString();
            nextBottom = new StringBuilder();
            p = 0;
        }
        if (bottom.length() == 1) return true;
        String key = bottom.substring(p,p + 2);
        if (map.containsKey(key)){
            for (String val : map.get(key)){
                nextBottom.append(val);
                if (dfs(map,bottom,nextBottom,p + 1)) return true;
                nextBottom.setLength(nextBottom.length() - 1);//还原
            }
        }
        return false;
    }
}