将字符串这章移动到前面，增加atoi等题目，修复morris遍历的BUG

C++/ACM-cheat-sheet.tex

@@ -37,6 +37,7 @@
 
 \include{chapProgrammingTrick}
 \include{chapLinearList}
+\include{chapString}
 \include{chapStackAndQueue}
 \include{chapTree}
 \include{chapSearching}
@@ -50,7 +51,6 @@
 \include{chapGraph}
 \include{chapMath}
 \include{chapBigInt}
-\include{chapString}
 \include{chapFundamental}
 
 \appendix % 开始附录，章用字母编号

C++/chapString.tex

@@ -1,13 +1,200 @@
 \chapter{字符串}
 
 
+\section{字符串API} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+面试中经常会出现，现场编写 \fn{strcpy, strlen, strstr, atoi} 等库函数的题目。这类题目看起来简单，实则难度很大，区分都很高，很容易考察出你的编程功底，是面试官的最爱。
+
+
+\subsection{strlen}
+
+
+\subsubsection{描述}
+实现 \fn{strlen}，获取字符串长度，函数原型如下：
+\begin{Code}
+size_t strlen(const char *str);
+\end{Code}
+
+
+\subsubsection{分析}
+
+
+
+\subsubsection{代码}
+\begin{Code}
+size_t strlen(const char *str) {
+    const char *s;
+    for (s = str; *s; ++s) {}
+    return(s - str);
+}
+\end{Code}
+
+
+\subsection{strcpy}
+
+
+\subsubsection{描述}
+实现 \fn{strcpy}，字符串拷贝函数，函数原型如下：
+\begin{Code}
+char* strcpy(char *to, const char *from);
+\end{Code}
+
+
+\subsubsection{分析}
+
+
+
+\subsubsection{代码}
+\begin{Code}
+char* strcpy(char *to, const char *from) {
+    assert(to != NULL && from != NULL);
+    char *p = to;
+    while ((*p++ = *from++) != '\0')
+        ;
+    return to;
+}
+\end{Code}
+
+
+\subsection{strstr}
+
+
+\subsubsection{描述}
+实现 \fn{strstr}，子串查找函数，函数原型如下：
+\begin{Code}
+char * strstr(const char *haystack, const char *needle);
+\end{Code}
+
+
+\subsubsection{分析}
+暴力算法的复杂度是 $O(m*n)$，代码如下。其他算法见第\S \ref{sec:substring-search}节 “子串查找”。
+
+
+\subsubsection{代码}
+\begin{Code}
+char *strstr(const char *haystack, const char *needle) {
+    // if needle is empty return the full string
+    if (!*needle) return (char*) haystack;
+
+    const char *p1;
+    const char *p2;
+    const char *p1_advance = haystack;
+    for (p2 = &needle[1]; *p2; ++p2) {
+        p1_advance++;   // advance p1_advance M-1 times
+    }
+
+    for (p1 = haystack; *p1_advance; p1_advance++) {
+        char *p1_old = (char*) p1;
+        p2 = needle;
+        while (*p1 && *p2 && *p1 == *p2) {
+            p1++;
+            p2++;
+        }
+        if (!*p2) return p1_old;
+
+        p1 = p1_old + 1;
+    }
+    return NULL;
+}
+\end{Code}
+
+
+\subsubsection{相关题目}
+与本题相同的题目：
+\begindot
+\item LeetCode Implement strStr(), \myurl{http://leetcode.com/oldoj\#question_28}
+\myenddot
+
+与本题相似的题目：
+\begindot
+\item  无
+\myenddot
+
+
+\subsection{atoi}
+\label{sec:string-to-integer}
+
+
+\subsubsection{描述}
+实现 \fn{atoi}，将一个字符串转化为整数，函数原型如下：
+\begin{Code}
+int atoi(const char *str);
+\end{Code}
+
+
+\subsubsection{分析}
+注意，这题是故意给很少的信息，让你来考虑所有可能的输入。
+
+来看一下\fn{atoi}的官方文档(\myurl{http://www.cplusplus.com/reference/cstdlib/atoi/})，看看它有什么特性：
+
+The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
+
+The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
+
+If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
+
+If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, \fn{INT_MAX (2147483647)} or \fn{INT_MIN (-2147483648)} is returned.
+
+注意几个测试用例：
+\begin{enumerate}
+\item 不规则输入，但是有效，"-3924x8fc"， "  +  413",
+\item 无效格式，" ++c", " ++1"
+\item 溢出数据，"2147483648"
+\end{enumerate}
+
+
+\subsubsection{代码}
+\begin{Code}
+int atoi(const char *str) {
+    int num = 0;
+    int sign = 1;
+    const int len = strlen(str);
+    int i = 0;
+
+    while (str[i] == ' ' && i < len) i++;
+
+    if (str[i] == '+') i++;
+
+    if (str[i] == '-') {
+        sign = -1;
+        i++;
+    }
+
+    for (; i < len; i++) {
+        if (str[i] < '0' || str[i] > '9')
+            break;
+        if (num > INT_MAX / 10 ||
+                (num == INT_MAX / 10 &&
+                        (str[i] - '0') > INT_MAX % 10)) {
+            return sign == -1 ? INT_MIN : INT_MAX;
+        }
+        num = num * 10 + str[i] - '0';
+    }
+    return num * sign;
+}
+\end{Code}
+
+
+\subsubsection{相关题目}
+与本题相同的题目：
+\begindot
+\item LeetCode String to Integer (atoi), \myurl{http://leetcode.com/oldoj\#question_8}
+\myenddot
+
+与本题相似的题目：
+\begindot
+\item  无
+\myenddot
+
+
 \section{字符串排序} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 
 \section{单词查找树} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 
 \section{子串查找} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\label{sec:substring-search}
+
 字符串的一种基本操作就是\textbf{子串查找}(substring search)：给定一个长度为$N$的文本和一个长度为$M$的模式串(pattern string)，在文本中找到一个与该模式相符的子字符串。
 
 最简单的算法是暴力查找，时间复杂度是$O(MN)$。下面介绍两个更高效的算法。

C++/chapTree.tex

@@ -425,7 +425,7 @@ \subsection{C语言实现}
 
             if (node->right == NULL ) { /* 还没线索化，则建立线索 */
                 node->right = cur;
-                prev = cur;
+                /* prev = cur; 不能有这句，cur还没有被访问 */
                 cur = cur->left;
             } else {    /* 已经线索化，则访问节点，并删除线索  */
                 visit(cur);
@@ -450,7 +450,7 @@ \subsection{C语言实现}
     while (cur != NULL ) {
         if (cur->left == NULL ) {
             visit(cur);
-            prev = cur;
+            prev = cur; /* cur刚刚被访问过 */
             cur = cur->right;
         } else {
             /* 查找前驱 */
@@ -459,13 +459,13 @@ \subsection{C语言实现}
                 node = node->right;
 
             if (node->right == NULL ) { /* 还没线索化，则建立线索 */
-                visit(cur); /* 仅这一行与中序不同 */
+                visit(cur); /* 仅这一行的位置与中序不同 */
                 node->right = cur;
-                prev = cur;
+                prev = cur; /* cur刚刚被访问过 */
                 cur = cur->left;
-            } else {    /* 已经线索化，则访问节点，并删除线索  */
+            } else {    /* 已经线索化，则删除线索  */
                 node->right = NULL;
-                prev = cur;
+                /* prev = cur; 不能有这句，cur已经被访问 */
                 cur = cur->right;
             }
         }
@@ -490,7 +490,7 @@ \subsection{C语言实现}
     cur = &dump;
     while (cur != NULL ) {
         if (cur->left == NULL ) {
-            prev = cur;
+            prev = cur; /* 必须要有 */
             cur = cur->right;
         } else {
             bt_node_t *node = cur->left;
@@ -499,12 +499,12 @@ \subsection{C语言实现}
 
             if (node->right == NULL ) { /* 还没线索化，则建立线索 */
                 node->right = cur;
-                prev = cur;
+                prev = cur; /* 必须要有 */
                 cur = cur->left;
             } else { /* 已经线索化，则访问节点，并删除线索  */
                 visit_reverse(cur->left, prev, visit);  // call print
                 prev->right = NULL;
-                prev = cur;
+                prev = cur; /* 必须要有 */
                 cur = cur->right;
             }
         }
@@ -539,7 +539,7 @@ \subsection{C语言实现}
         int (*visit)(bt_node_t*)) {
     bt_node_t *p = to;
     reverse(from, to);
-    
+
     while (1) {
         visit(p);
         if (p == from)

C/ACM-cheat-sheet.tex

@@ -37,6 +37,7 @@
 
 \include{chapProgrammingTrick}
 \include{chapLinearList}
+\include{chapString}
 \include{chapStackAndQueue}
 \include{chapTree}
 \include{chapSearching}
@@ -50,7 +51,6 @@
 \include{chapGraph}
 \include{chapMath}
 \include{chapBigInt}
-\include{chapString}
 \include{chapFundamental}
 
 \appendix % 开始附录，章用字母编号