标签：cas tput sort you letter letters mon val 单词
There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.
Example 1:
Input:
[
"wrt",
"wrf",
"er",
"ett",
"rftt"
]
Output: "wertf"
Example 2:
Input:
[
"z",
"x"
]
Output: "zx"
Example 3:
Input: [ "z", "x", "z" ] Output:""Explanation: The order is invalid, so return"".
Note:
这题看题目就知道是拓扑排序解，但是怎么构建图，以及一些不是字典序的输入要判断，有些corner case要考虑到。
首先输入有可能不是拓扑排序的，一个是有环，如例3，这个在走完拓扑的顺序，如果输出结点和原结点数目不一致，可以判断。另外一种可能是短词排在后面如：[‘err‘, ‘e‘] 这种单纯通过拓扑排序看不出来，所以需要在建图时判断。另外注意我们只是针对有边的情况建图，很多时候有结点无边，所以建完的图key值不一定能完全覆盖结点。所以把所有字母组成的结点单求出来很重要。另外注意组合只有一个单词的情况。代码如下：
class Solution(object): def alienOrder(self, words): """ :type words: List[str] :rtype: str """ # 可能有环， 每个单词只有一个字符的情况, 不符合字典序的情况，[‘err‘, ‘e‘] graph = collections.defaultdict(set) nodes = set() nodes.update(words[0]) for i in xrange(1, len(words)): diff = 0 for j in xrange(min(len(words[i-1]), len(words[i]))): if words[i-1][j] != words[i][j]: graph[words[i-1][j]].add(words[i][j]) diff += 1 break if diff == 0 and len(words[i-1]) > len(words[i]): return "" nodes.update(words[i]) hashmap = collections.defaultdict(int) for node in nodes: for n in graph[node]: hashmap[n] += 1 queue = collections.deque() for node in graph: if node not in hashmap: queue.append(node) res = [] while queue: node = queue.popleft() res.append(node) for n in graph[node]: hashmap[n] -= 1 if hashmap[n] == 0: queue.append(n)
#判断是否有环 if len(res) != len(nodes): res = [] return "".join(res)
标签：cas tput sort you letter letters mon val 单词
原文地址：https://www.cnblogs.com/sherylwang/p/9745158.html
