面试笔试杂项积累-leetcode 331-335

至此2016-2-15，全部333道题除去锁题和不想做的= =。就都做完了。。。

331.331-Verify Preorder Serialization of a Binary Tree-Difficulty: Medium

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as#.

     _9_    /   \   3     2  / \   / \ 4   1  #  6/ \ / \   / \# # # #   # #

For example, the above binary tree can be serialized to the string"9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character'#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as"1,,3".

Example 1:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

思路

判断先序遍历能否组成一棵树

运用正则表达式，然后一点一点“砍叶”，因为不全的地方包括左右子节点都会用#代替，那么1-9，#，#就是一个叶节点，我们把尾端每个叶节点都用#代替，下面一层叶节点就没了，反复几次，知道没有1-9，#，#叶节点了，如果最后剩的是#空的根，就说明节点全部砍光，就是一个二叉树，否则不是

public class Solution {    public bool IsValidSerialization(string preorder) {        var t = String.Empty;        preorder = System.Text.RegularExpressions.Regex.Replace(preorder, "[0-9]+", "0");        while (preorder != t)        {            t = preorder;            preorder = preorder.Replace("0,#,#", "#");        }        return preorder == "#";    }}

332.332-Reconstruct Itinerary-Difficulty: Medium

Given a list of airline tickets represented by pairs of departure and arrival airports[from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs fromJFK. Thus, the itinerary must begin with JFK.

Note:

1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets form at least one valid itinerary.

Example 1:
tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:
tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

思路

最简路径

博主开始的代码可以求出最简，但是顺序不对，难道这个不是有多种可能性的吗。。。不知道原因
参考：

https://leetcode.com/discuss/86271/c%23-recursive-solution

public class Solution {    private IDictionary<string, List<string>> m_routes;    private int m_count;    public IList<string> FindItinerary(string[,] tickets)    {        m_routes = new Dictionary<string, List<string>>();        m_count = tickets.GetLength(0);        for (int i = 0; i < m_count; i++)        {            var s = tickets[i, 0];            var d = tickets[i, 1];            if (!m_routes.ContainsKey(s))                m_routes.Add(s, new List<string>());            m_routes[s].Add(d);        }        foreach (var list in m_routes.Values)            list.Sort();        return GetPath("JFK");    }    private IList<string> GetPath(string s)    {        if (m_routes.ContainsKey(s) && m_routes[s].Count > 0)        {            var children = m_routes[s];            for (int i = 0; i < children.Count; i++)            {                var child = children[i];                children.RemoveAt(i);                m_count--;                var best = GetPath(child);                children.Insert(i, child);                m_count++;                if (best != null)                {                    best.Insert(0, s);                    return best;                }            }            return null;        }        if (m_count == 0)            return new List<string> {s};        return null;    }}