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:
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'}
.
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;
}
}