kmp

he3210 · Dec 7, 2021 · b2f3981 · b2f3981
1 parent b530cf0
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diff --git a/algorithm notebook/.#二分查找.org b/algorithm notebook/.#二分查找.org
@@ -0,0 +1 @@
+[email protected]
diff --git a/algorithm notebook/algorithm.org b/algorithm notebook/algorithm.org
@@ -59,3 +59,4 @@
  - [[./二叉排序树.org][二叉排序树]]
  - [[./平衡二叉树.org][平衡二叉树]]
  - [[./红黑树.org][红黑树]]
+ - 线段树
diff --git a/algorithm notebook/kmp.org b/algorithm notebook/kmp.org
@@ -7,124 +7,92 @@
 * BF 解法
 
 #+BEGIN_SRC c
-#include<assert.h>
-int BF(const char *str, const char *substr)
-{
-    assert(str && substr);
-    int lenStr = strlen(str);
-    int lenSubstr = strlen(substr);
-    if(lenStr < lenSubstr)
-        return -1;
-
-    for(int i = 0; i != lenStr-lenSubstr+1; ++i)
-    {
-        int j;
-        for(j = 0; j != lenSubstr; ++j)
-            if(str[i+j] != substr[j])
-                break;
-        if(j == lenSubstr)
-            return i;
-    }
-    return -1;
-}
+  int BF(const char *str, const char *substr) {
+      assert(str && substr);
+      int n = strlen(str);
+      int m = strlen(substr);
+      for(int i = 0; i < n - m + 1; ++i) {
+          int j;
+          for(j = 0; j < m; ++j)
+              if(str[i+j] != substr[j])
+                  break;
+          if(j == m)
+              return i;
+      }
+      return -1;
+  }
 #+END_SRC
 
 * KMP 解法
 #+BEGIN_SRC c
   #include<stdio.h>
   #include<string.h>
-
-  // 获取指针数组 previous
-  void getPrevious(const char *pattern, int lenPattern, int *previous)
-  {
-      int curPos = 0;
-      previous[curPos] = -1;
-      int flagPos = -1;
-      while(curPos < lenPattern-1)
-      {
-          if(-1 == flagPos || pattern[curPos] == pattern[flagPos])
-          {
-              ++curPos;
-              ++flagPos;
-              previous[curPos] = flagPos;
+  #include<stdlib.h>
+
+  void get_next(const char *pattern, int plen, int *next) {
+      int i = 0;
+      next[i] = -1;  // 用来保存主串和模式串失配时，模式串应向前回溯的位置
+      int j = -1;
+      while(i < plen-1) {
+          if(j == -1 || pattern[i] == pattern[j]) {  // 例如 "xyxa" 中的两个 'x'
+              ++i;
+              ++j;
+              next[i] = j;  // 例如 "xyxa"，主串和 'a' 失配后，应该回溯到 'y'，主串再和 'y' 比较一下
+          } else {
+              j = next[j];
           }
-          else
-              flagPos = previous[flagPos];
       }
   }
 
-  int KMP(const char *str, int lenStr, const char *pattern, int lenPattern, const int *previous, int pos)
-  {
-      int strPos = pos;
-      int patternPos = 0;
-      while(strPos != lenStr && patternPos != lenPattern)
-      {
-          if(-1 == patternPos || str[strPos] == pattern[patternPos])
-          {
-              ++strPos;
-              ++patternPos;
+  // 改进版本
+  void get_next_v2(const char *pattern, int plen, int *next) {
+      int i = 0;
+      next[i] = -1;
+      int j = -1;
+      while(i < plen-1) {
+          if(j == -1 || pattern[i] == pattern[j]) {  // 模式串中位置连续的两个子串，上一个字符匹配上了。例如 "xyxy" 中下标 0 的 'x' 和下标 2 的 'x'
+              ++i;
+              ++j;
+              if(pattern[i] == pattern[j])  // 模式串中的两个子串的下一个字符也匹配上了，例如 "xyxy" 的子串 "xy" "xy" 中 'y'。如果主串和模式串下标 3 的 'y' 失配的话，肯定也和下标 1 的 'y' 失配，所以可以向前回溯两次
+                  next[i] = next[j];
+              else
+                  next[i] = j;  // 例如 "xyxa" 中的 'y' 和 'a'。如果主串和模式串中 'a' 失配的话，'a' 又和 'y' 不相等，所以主串不一定和 'y' 失配，所以应该回溯到 'y'
+          } else {
+              j = next[j];
           }
-          else
-              patternPos = previous[patternPos];
       }
-      if(patternPos == lenPattern)
-          return strPos-lenPattern;
-      else
-          return -1;
   }
 
-  int main(void)
-  {
-      const char *str = "How are you";
-      const char *substr = "are";
-      printf("%d\n", BF(str, substr));
-
-      const char *str2 = "001012012301234";
-      const char *substr2 = "0120123";
-      int previous[1024];
-      getPrevious(substr2, strlen(substr2), previous);
-      printf("%d\n", KMP(str2, strlen(str2), substr2, strlen(substr2), previous, 0));
+  int KMP(const char *str, int slen, const char *pattern, int plen, const int *next) {
+      int i = 0, j = 0;
+      while(i < slen && j < plen) {
+          if(j == -1 || str[i] == pattern[j]) {
+              ++i;
+              ++j;
+          } else {
+              j = next[j];
+          }
+      }
+      if(j == plen)
+          return i - plen;
+      return -1;
+  }
 
-      return 0;
+  int main(int argc, char **argv) {
+      char str[] = "abxyxyxyyxxyxadf";
+      char pattern[] = "xyxyyxxyx";
+      int slen = strlen(str);
+      int plen = strlen(pattern);
+      int *next = (int*)calloc(plen, sizeof(int));
+      get_next_v2(pattern, plen, next);
+      printf("%d\n", KMP(str, slen, pattern, plen, next));
   }
 #+END_SRC
 
-* 优化
-
-#+BEGIN_SRC c
-void getPrevious(const char *pattern, int lenPattern, int *previous)
-{
-    int curPos = 0;
-    previous[curPos] = -1;
-    int flagPos = -1;
-    while(curPos < lenPattern-1)
-    {
-        if(-1 == flagPos || pattern[curPos] == pattern[flagPos])
-        {
-            ++curPos;
-            ++flagPos;
-            if(pattern[curPos] != pattern[flagPos])
-                previous[curPos] = flagPos;
-            else
-                /* 此时 pattern[curPos-1] 和 pattern[flagPos-1] 相等、
-                   pattern[curPos] 和 pattern[flagPos] 相等。
-                   这样的字符串形式为："...xy...xy...",
-                   而 previous 数组存在的意义是源串和模式串不匹配时模式串应该回溯的下标
-                   第一个 "xy" 和第二个 "xy" 相同，由于源串和第二个 "xy" 不匹配，所以肯定和第一个 "xy" 也不匹配（例如：源串为 xyxx，模式串为 xyxy）
-                   故向前回溯时，应跳过第一个 "xy"，使用 previous[curPos] = previous[flagPos] 语句
-                   这样生成的 previous 数组可以使 KMP 函数中少一些不必要的比较
-                 */
-                previous[curPos] = previous[flagPos];
-        }
-        else
-            flagPos = previous[flagPos];
-    }
-}
-#+END_SRC
 
-* previous 数组
+* next 数组
 
-对于模式串"xyxyyxxyx"，previous数组为
+对于模式串"xyxyyxxyx"，next 数组为
 | 下标  |  0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
 |-------+----+---+---+---+---+---+---+---+---|
 | value | -1 | 0 | 0 | 1 | 2 | 0 | 1 | 1 | 2 |

diff --git a/images/jetbrains-agent.jar.zip b/images/jetbrains-agent.jar.zip
diff --git a/images/个人收入证明.pdf b/images/个人收入证明.pdf
diff --git a/images/时中贺_应聘后台开发岗.pdf b/images/时中贺_应聘后台开发岗.pdf
diff --git a/images/离职证明.pdf b/images/离职证明.pdf
diff --git a/qu/clion 注册码.org b/qu/clion 注册码.org
@@ -1,46 +1,5 @@
 #+TITLE: clion 注册码
 
-* 方法一
-#+BEGIN_QUOTE
-点击下方链接进去下载(普通下载)
-
-1月激活码---
-https://www.lanzous.com/b0ev9wx1g?
-密码:77ab
-
-激活码必须下载到电脑在复制，不然失效，如何激活：软件菜单-帮助-注册，把激活码全部复制到软件激活窗口中的Activation code
-
-激活不成功 请看这里：?https://shimo.im/docs/8icBv925HHESLZ1w/?
-感谢您的光临~
-
-或者使用以下九个账号之一  密码都是Aa123456
-2020141
-2020142
-2020143
-2020144
-2020145
-2020146
-2020147
-2020148
-2020149
-#+END_QUOTE
-
-* 方法二
-[[../images/jetbrains-agent.jar.zip][jetbrains-agent.jar.zip]]
-
-- 点击试用进入 IDE
-- 点击 Help -> Edit Custom VM Options ...
-- 在打开的vmoptions编辑窗口末行添加 jetbrains-agent.jar 文件的绝对路径
- #+BEGIN_QUOTE
- -javaagent:/Users/he/Downloads/crack_file/jetbrains-agent/jetbrains-agent.jar
- #+END_QUOTE
-- 重启 IDE。点击 Help -> Register...
-- 选择License server方式，地址填入：http://fls.jetbrains-agent.com
-- 点击激活
-* 激活码
-#+BEGIN_QUOTE
-3AGXEJXFK9-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-WGTHs6XpDhr+uumvbwQPOdlxWnQwgnGaL4eRnlpGKApEEkJyYvNEuPWBSrQkPmVpim/8Sab6HV04Dw3IzkJT0yTc29sPEXBf69+7y6Jv718FaJu4MWfsAk/ZGtNIUOczUQ0iGKKnSSsfQ/3UoMv0q/yJcfvj+me5Zd/gfaisCCMUaGjB/lWIPpEPzblDtVJbRexB1MALrLCEoDv3ujcPAZ7xWb54DiZwjYhQvQ+CvpNNF2jeTku7lbm5v+BoDsdeRq7YBt9ANLUKPr2DahcaZ4gctpHZXhG96IyKx232jYq9jQrFDbQMtVr3E+GsCekMEWSD//dLT+HuZdc1sAIYrw==-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
-#+END_QUOTE
 * 2021.02.25已激活clion
 - 链接 https://tech.souyunku.com/?p=16189
 - 修改 reset_jetbrains_eval_mac_linux.sh 脚本变量 JB_PRODUCTS，并执行