Updated chapter String

soulmachine · soulmachine · commit 7611b95d3ae6 · 2016-01-10T12:37:48.000-08:00
diff --git a/C++/chapString.tex b/C++/chapString.tex
@@ -35,7 +35,7 @@ \subsubsection{代码}
             if (!::isalnum(*left))  ++left;
             else if (!::isalnum(*right)) --right;
             else if (*left != *right) return false;
-            else{ left++, right--; }
+            else { left++, right--; }
         }
         return true;
     }
@@ -69,29 +69,20 @@ \subsubsection{暴力匹配}
 // 暴力解法，时间复杂度O(N*M)，空间复杂度O(1)
 class Solution {
 public:
-    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++;
+    int strStr(const string& haystack, const string& needle) {
+        if (needle.empty()) return 0;
+
+        const int N = haystack.size() - needle.size() + 1;
+        for (int i = 0; i < N; i++) {
+            int j = i;
+            int k = 0;
+            while (j < haystack.size() && k < needle.size() && haystack[j] == needle[k]) {
+                j++;
+                k++;
             }
-            if (!*p2) return p1_old;
-
-            p1 = p1_old + 1;
+            if (k == needle.size()) return i;
         }
-        return nullptr;
+        return -1;
     }
 };
 \end{Code}
@@ -103,10 +94,8 @@ \subsubsection{KMP}
 // KMP，时间复杂度O(N+M)，空间复杂度O(M)
 class Solution {
 public:
-    char *strStr(const char *haystack, const char *needle) {
-        int pos = kmp(haystack, needle);
-        if (pos == -1) return nullptr;
-        else return (char*)haystack + pos;
+    int strStr(const string& haystack, const string& needle) {
+        return kmp(haystack.c_str(), needle.c_str());
     }
 private:
     /*
@@ -389,7 +378,7 @@ \subsubsection{动规}
 // 动规，时间复杂度O(n^2)，空间复杂度O(n^2)
 class Solution {
 public:
-    string longestPalindrome(string s) {
+    string longestPalindrome(const string& s) {
         const int n = s.size();
         bool f[n][n];
         fill_n(&f[0][0], n * n, false);
@@ -423,7 +412,7 @@ \subsubsection{Manacher’s Algorithm}
     // Transform S into T.
     // For example, S = "abba", T = "^#a#b#b#a#$".
     // ^ and $ signs are sentinels appended to each end to avoid bounds checking
-    string preProcess(string s) {
+    string preProcess(const string& s) {
         int n = s.length();
         if (n == 0) return "^$";
 
@@ -520,6 +509,10 @@ \subsubsection{递归版}
 // 递归版，时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
+    bool isMatch(const string& s, const string& p) {
+        return isMatch(s.c_str(), p.c_str());
+    }
+private:
     bool isMatch(const char *s, const char *p) {
         if (*p == '\0') return *s == '\0';
 
@@ -596,6 +589,10 @@ \subsubsection{递归版}
 // 时间复杂度O(n!*m!)，空间复杂度O(n)
 class Solution {
 public:
+    bool isMatch(const string& s, const string& p) {
+        return isMatch(s.c_str(), p.c_str());
+    }
+private:
     bool isMatch(const char *s, const char *p) {
         if (*p == '*') {
             while (*p == '*') ++p;  //skip continuous '*'
@@ -618,6 +615,10 @@ \subsubsection{迭代版}
 // 迭代版，时间复杂度O(n*m)，空间复杂度O(1)
 class Solution {
 public:
+    bool isMatch(const string& s, const string& p) {
+        return isMatch(s.c_str(), p.c_str());
+    }
+private:
     bool isMatch(const char *s, const char *p) {
         bool star = false;
         const char *str, *ptr;
@@ -751,7 +752,7 @@ \subsubsection{有限自动机}
 // finite automata，时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
-    bool isNumber(const char *s) {
+    bool isNumber(const string& s) {
         enum InputType {
             INVALID,    // 0
             SPACE,      // 1
@@ -774,17 +775,17 @@ \subsubsection{有限自动机}
                 };
 
         int state = 0;
-        for (; *s != '\0'; ++s) {
+        for (auto ch : s) {
             InputType inputType = INVALID;
-            if (isspace(*s))
+            if (isspace(ch))
                 inputType = SPACE;
-            else if (*s == '+' || *s == '-')
+            else if (ch == '+' || ch == '-')
                 inputType = SIGN;
-            else if (isdigit(*s))
+            else if (isdigit(ch))
                 inputType = DIGIT;
-            else if (*s == '.')
+            else if (ch == '.')
                 inputType = DOT;
-            else if (*s == 'e' || *s == 'E')
+            else if (ch == 'e' || ch == 'E')
                 inputType = EXPONENT;
 
             // Get next state from current state and input symbol
@@ -809,6 +810,10 @@ \subsubsection{使用strtod()}
 // 偷懒，直接用 strtod()，时间复杂度O(n)
 class Solution {
 public:
+    bool isNumber (const string& s) {
+        return isNumber(s.c_str());
+    }
+private:
     bool isNumber (char const* s) {
         char* endptr;
         strtod (s, &endptr);
@@ -909,7 +914,7 @@ \subsubsection{代码}
         }
     }
 
-    int romanToInt(string s) {
+    int romanToInt(const string& s) {
         int result = 0;
         for (size_t i = 0; i < s.size(); i++) {
             if (i > 0 && map(s[i]) > map(s[i - 1])) {
@@ -1070,7 +1075,7 @@ \subsubsection{代码}
 // 时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
-    string simplifyPath(string const& path) {
+    string simplifyPath(const string& path) {
         vector<string> dirs; // 当做栈
 
         for (auto i = path.begin(); i != path.end();) {
@@ -1137,10 +1142,9 @@ \subsubsection{用 STL}
 // 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
-    int lengthOfLastWord(const char *s) {
-        const string str(s);
-        auto first = find_if(str.rbegin(), str.rend(), ::isalpha);
-        auto last = find_if_not(first, str.rend(), ::isalpha);
+    int lengthOfLastWord(const string& s) {
+        auto first = find_if(s.rbegin(), s.rend(), ::isalpha);
+        auto last = find_if_not(first, s.rend(), ::isalpha);
         return distance(first, last);
     }
 };
@@ -1154,6 +1158,10 @@ \subsubsection{顺序扫描}
 // 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
+    int lengthOfLastWord(const string& s) {
+        return lengthOfLastWord(s.c_str());
+    }
+private:
     int lengthOfLastWord(const char *s) {
         int len = 0;
         while (*s) {
