[leetcode/lintcode 题解] 算法面试高频题：序列重构

描述
判断是否序列 org 能唯一地由 seqs重构得出. org是一个由从1到n的正整数排列而成的序列，1≤n≤104。 重构表示组合成seqs的一个最短的父序列 (意思是，一个最短的序列使得所有 seqs里的序列都是它的子序列).
判断是否有且仅有一个能从 seqs重构出来的序列，并且这个序列是org。

在线评测地址：领扣题库官网

样例1
输入:org = [1,2,3], seqs = [[1,2],[1,3]]
输出: false
解释:
[1,2,3] 并不是唯一可以被重构出的序列，还可以重构出 [1,3,2]

样例2
输入: org = [1,2,3], seqs = [[1,2]]
输出: false
解释:
能重构出的序列只有 [1,2].

样例3
输入: org = [1,2,3], seqs = [[1,2],[1,3],[2,3]]
输出: true
解释:
序列 [1,2], [1,3], 和 [2,3] 可以唯一重构出 [1,2,3].

样例4
输入:org = [4,1,5,2,6,3], seqs = [[5,2,6,3],[4,1,5,2]]
输出:true

题解
九章算法班里讲过的拓扑排序，只要保证 queue 里最多同时只有一个元素即可。 所以这是 queue 用 list 然后每次 pop 也可以，反正只有一个数。

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):
        graph = self.build_graph(seqs)
        topo_order = self.topological_sort(graph)
        return topo_order == org
            
    def build_graph(self, seqs):
        # initialize graph
        graph = {}
        for seq in seqs:
            for node in seq:
                if node not in graph:
                    graph[node] = set()
        
        for seq in seqs:
            for i in range(1, len(seq)):
                graph[seq[i - 1]].add(seq[i])

        return graph
    
    def get_indegrees(self, graph):
        indegrees = {
            node: 0
            for node in graph
        }
        
        for node in graph:
            for neighbor in graph[node]:
                indegrees[neighbor] += 1
                
        return indegrees
        
    def topological_sort(self, graph):
        indegrees = self.get_indegrees(graph)
        
        queue = []
        for node in graph:
            if indegrees[node] == 0:
                queue.append(node)
        
        topo_order = []
        while queue:
            if len(queue) > 1:
                # there must exist more than one topo orders
                return None
                
            node = queue.pop()
            topo_order.append(node)
            for neighbor in graph[node]:
                indegrees[neighbor] -= 1
                if indegrees[neighbor] == 0:
                    queue.append(neighbor)
                    
        if len(topo_order) == len(graph):
            return topo_order
            
        return None

更多题解参考：九章官网solution