Lintcode 605. Sequence Reconstruction

又是一道拓扑排序的变形题, Given org = [1,2,3], seqs = [[1,2],[1,3]]
给一串序列看后面能否组成这个序列。
思路：拓扑排序的题就要记录邻接矩阵，然后和每个点的入度。入度为0放入队列，然后取出队列里的值，让它的邻居入度都减1，如果入度为0的点继续放在队列里，依次循环。因为拓扑排序是任意输出一组结果，但这里要求结果是唯一的。要求不能同时出现两个入度为0的点放入队列。所以我们可以设计一个level值来记录bfs的层数。有点像树的层级输出，并且分层放置。这里面还得加一个node的set， 来存放seqs里面的所有node，最后判断true or false就是判断是否生成的序列和原序列一致，是否node的个数和原序列一致，level长度和原序列是否一致。
python代码：

class Solution:
    """
    @param org: a permutation of the integers from 1 to n
    @param seqs: a list of sequences
    @return: true if it can be reconstructed only one or false
    """
    
    def sequenceReconstruction(self, org, seqs):
        # write your code here
        from collections import deque
        neighbor = {}
        indegree = {}
        nodes = set()
        
        for i in org:
            neighbor[i] = []
            indegree[i] = 0
            
        for pair in seqs:
            nodes = nodes | set(pair)
            for i in range(len(pair) - 1):
                if pair[i] not in neighbor or pair[i+1] not in neighbor:
                    return False
                neighbor[pair[i]].append(pair[i+1])
                indegree[pair[i+1]] += 1
        
        queue = deque()
        level = 0
        res = []
        for key in indegree:
            if indegree[key] == 0:
                queue.append(key)
                
        
        
        while queue:
            size = len(queue)
            level += 1
            for _ in range(size):
                node = queue.popleft()
                res.append(node)
                for i in neighbor[node]:
                    indegree[i] -= 1
                    if indegree[i] == 0:
                        queue.append(i)
        
        return res == org and len(nodes) == len(org) and level == len(org)