添加 Longest Consecutive Sequence, Valid Palindrome，以及王顺达的一个代码

Author: soulmachine

C++/LeetCodet题解(C++版).pdf

@@ -286,6 +286,7 @@ \subsubsection{代码}
 
 \begin{Code}
 // LeetCode, Multiply Strings
+// @author 连城(http://weibo.com/lianchengzju)
 // 一个字符对应一个int
 typedef vector<int> bigint;
 

C++/chapImplement.tex

@@ -314,6 +314,72 @@ \subsubsection{相关题目}
 \myenddot
 
 
+\subsection{Longest Consecutive Sequence} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\label{sec:longest-consecutive-sequence}
+
+
+\subsubsection{描述}
+Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
+
+For example,
+Given \code{[100, 4, 200, 1, 3, 2]},
+The longest consecutive elements sequence is \code{[1, 2, 3, 4]}. Return its length: 4.
+
+Your algorithm should run in $O(n)$ complexity.
+
+
+\subsubsection{分析}
+如果允许$O(n \log n)$的复杂度，那么可以先排序，可是本题要求$O(n)$。
+
+由于序列里的元素是无序的，又要求$O(n)$，首先要想到用哈希表。
+
+用一个哈希表 \fn{unordered_map<int, bool> used}记录每个元素是否使用，对每个元素，以该元素为中心，往左右扩张，直到不连续为止，记录下最长的长度。
+
+
+\subsubsection{代码}
+\begin{Code}
+// Leet Code, Longest Consecutive Sequence
+class Solution {
+public:
+    int longestConsecutive(vector<int> const& num) {
+        unordered_map<int, bool> used;
+
+        for (auto i : num) used[i] = false;
+
+        int longest = 0;
+
+        for (auto i : num) {
+            if (used[i]) continue;
+
+            int length = 1;
+
+            used[i] = true;
+
+            for (int j = i + 1; used.find(j) != used.end(); ++j) {
+                used[j] = true;
+                ++length;
+            }
+
+            for (int j = i - 1; used.find(j) != used.end(); --j) {
+                used[j] = true;
+                ++length;
+            }
+
+            longest = max(longest, length);
+        }
+
+        return longest;
+    }
+};
+\end{Code}
+
+
+\subsubsection{相关题目}
+\begindot
+\item 无
+\myenddot
+
+
 \section{单链表} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 单链表节点的定义如下：

C++/chapLinearList.tex

@@ -1,6 +1,63 @@
 \chapter{字符串}
 
 
+\section{Valid Palindrome} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\label{sec:valid-palindrome}
+
+
+\subsubsection{描述}
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+For example,\\
+\code{"A man, a plan, a canal: Panama"} is a palindrome.\\
+\code{"race a car"} is not a palindrome.
+
+Note:
+Have you consider that the string might be empty? This is a good question to ask during an interview.
+
+For the purpose of this problem, we define empty string as valid palindrome.
+
+
+\subsubsection{分析}
+无
+
+
+\subsubsection{代码}
+\begin{Code}
+// Leet Code, Valid Palindrome
+class Solution {
+public:
+    bool isPalindrome(string s) {
+        transform(s.begin(), s.end(), s.begin(), ::tolower);
+        auto left = s.begin(), right = prev(s.end());
+        while (left < right) {
+            if (!::isalnum(*left)) {
+                ++left;
+                continue;
+            }
+            if (!::isalnum(*right)) {
+                --right;
+                continue;
+            }
+            if (*left == *right) {
+                ++left;
+                --right;
+            } else {
+                return false;
+            }
+        }
+        return true;
+    }
+};
+\end{Code}
+
+
+\subsubsection{相关题目}
+\begindot
+\item 无
+\myenddot
+
+
 \section{Implement strStr()} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \label{sec:strstr}
 

C++/chapString.tex

@@ -656,7 +656,28 @@ \subsubsection{分析}
 \subsubsection{代码}
 递归版
 \begin{Code}
-// LeetCode, Flatten Binary Tree to Linked List，递归版
+// LeetCode, Flatten Binary Tree to Linked List
+// 递归版1，作者：王顺达，http://weibo.com/u/1234984145
+class Solution {
+public:
+    void flatten(TreeNode *root) {
+        flatten(root, NULL);
+    }
+private:
+    // 把root所代表树变成链表后，tail跟在该链表后面
+    TreeNode *flatten(TreeNode *root, TreeNode *tail) {
+        if (NULL == root) return tail;
+
+        root->right = flatten(root->left, flatten(root->right, tail));
+        root->left = NULL;
+        return root;
+    }
+};
+\end{Code}
+
+\begin{Code}
+// LeetCode, Flatten Binary Tree to Linked List
+// 递归版2
 class Solution {
 public:
     void flatten(TreeNode *root) {