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  • 面试宝典:字符串基础

    内容来自:面试宝典 字符串章节

    • 字符串与数字的相互转换
    • 实现strcopy
    • 求一个字符串中连续出现次数最多的子串
    • 求一个字符串中相同且最长的子串
    • 实现strstr
    • 将一句话中的单词进行倒置,标点符号不变。比如一句话:i come from tiamjin. 倒换后称为 tianjin. from come i
    • 计算从1-n之间的中包含1个个数
    • 转变字符串格式为原来字符串里的字符+该字符出现的次数的个数
      1 /*
      2  * mystring.h
      3  *
      4  *  Created on: Mar 10, 2017
      5  *      Author: wxquare
      6  */
      7 
      8 #ifndef MYSTRING_H_
      9 #define MYSTRING_H_
     10 
     11 #include <cassert>
     12 #include <string>
     13 #include <iostream>
     14 
     15 namespace MYSTRING {
     16 
     17 void itoa(int val, char* str) {
     18 
     19     char* tmp = new char[10];
     20 
     21     int i = 0;
     22     while (val) {
     23         tmp[i++] = (val % 10) + '0';
     24         val = val / 10;
     25     }
     26     i = i - 1;
     27     int j = 0;
     28     while (i >= 0) {
     29         str[j++] = tmp[i--];
     30     }
     31     delete[] tmp;
     32 }
     33 
     34 int atoi(char* str) {
     35 
     36     int i = 0, res = 0;
     37     while (str[i] != '') {
     38         res = res * 10 + str[i++] - '0';
     39     }
     40     return res;
     41 }
     42 
     43 char* strcopy(char* strDest, const char* strSrc) {
     44 
     45     assert(strDest != nullptr && strSrc != nullptr);
     46 
     47     char* p = strDest;
     48     char* q = const_cast<char*>(strSrc);
     49     while (*q != '') {
     50         *p++ = *q++;
     51     }
     52     *p = '';  //notice
     53     return strDest;
     54 }
     55 
     56 /*
     57  * 求一个字符串中连续出现次数最多的字串
     58  * maximum number of consecutive strings in a string.
     59  */
     60 std::string maxCounSubStr(const std::string &str) {
     61     int len = str.length();
     62     std::string res = str.substr(0, 1);
     63     int maxCount = 1, count;
     64     for (int i = 0; i < len; i++) {
     65         for (int j = i + 1; j < len; j++) {
     66             int offset = j - i;
     67             if (str.substr(i, offset) == str.substr(j, offset)) {
     68                 count = 2;
     69                 for (int k = j + offset; k < len; k += offset) {
     70                     if (str.substr(i, offset) == str.substr(k, offset)) {
     71                         count++;
     72                     } else {
     73                         break;
     74                     }
     75                 }
     76                 if (count > maxCount) {
     77                     maxCount = count;
     78                     res = str.substr(i, offset);
     79                 }
     80             }
     81         }
     82     }
     83     return res;
     84 }
     85 
     86 /*
     87  * 求一个字符串中相同且最长的字串
     88  */
     89 using string = std::string;
     90 string sameAndLongestSubstr(const string& str) {
     91     string res;
     92     int len = str.length();
     93     bool isfound = false;
     94     for (int l = len; l >= 1; l--) {
     95         if (isfound)
     96             break;
     97         for (int i = 0; i < len; i++) {
     98             if (i + l < len) {
     99                 string subStr = str.substr(i, l);
    100                 if (str.find(subStr) != str.rfind(subStr)) {
    101                     res = subStr;
    102 //                    std::cout << subStr << '	' << i << std::endl;
    103                     isfound = true;
    104                     break;
    105                 }
    106             }
    107         }
    108     }
    109     return res;
    110 }
    111 
    112 /*
    113  * implement strstr
    114  * char * strstr ( const char * str1, const char * str2 );
    115  */
    116 char* strstr(const char* str1, const char* str2) {
    117     char *cur = const_cast<char*>(str1);
    118     char *p1, *p2;
    119     while (cur != '') {
    120         p1 = cur;
    121         p2 = const_cast<char*>(str2);
    122         if (*p1 == *p2) {
    123             while (*p2 != '' && *p1 != '' && *p1++ == *p2++)
    124                 ;
    125             if (*p2 == '')
    126                 return cur;
    127             else if (*p1 == '')
    128                 return NULL;
    129             else {
    130                 cur++;
    131             }
    132         } else {
    133             cur++;
    134         }
    135     }
    136     return NULL;
    137 }
    138 
    139 /*
    140  * 把一句话里面的单词进行倒置
    141  * eg. "i come from tianjin."
    142  *        "tianjin. from come i"
    143  */
    144 void reverseWords(string& sentence) {
    145     int len = sentence.length();
    146     char* tmp = new char[len];
    147 
    148     //整体倒置
    149     int i = len - 1, j = 0;
    150     while (i >= 0) {
    151         tmp[j++] = sentence[i--];
    152     }
    153 
    154     //每个单词的逆序
    155     int k = 0;
    156     i = 0, j = 0;
    157     while (i < len) {
    158         for (int j = i; j < len; j++) {
    159             if (tmp[j] == ' ' || j == len - 1) {
    160                 if (j != len - 1) {
    161                     int n = j - i;
    162                     i = j + 1;
    163                     while (n--) {
    164                         sentence[k++] = tmp[--j];
    165                     }
    166                     sentence[k++] = ' ';
    167                 } else {
    168                     int n = j - i + 1;
    169                     i = j + 1;
    170                     while (n--) {
    171                         sentence[k++] = tmp[j--];
    172                     }
    173                 }
    174                 break;
    175             }
    176         }
    177     }
    178     delete [] tmp;
    179 }
    180 
    181 
    182 /*
    183  * 统计1的个数
    184  * ege: f(13) = 6  {1,2,3,4,5,6,7,8,9,10,11,12,13}
    185  *
    186  */
    187 int countOne(int n){
    188     if(n == 1) return 1;
    189     int r = n;
    190     int count = 0;
    191     while(r){
    192         if(r % 10 == 1){
    193             count++;
    194         }
    195         r = r / 10;
    196     }
    197     return countOne(n-1) + count;
    198 }
    199 
    200 
    201 /*
    202  * 把字符串转为字符加上数字的表述方式
    203  * ege: 1233444  转变为 11213243
    204  *
    205  */
    206 string CharNumber(const string& str){
    207     string res;
    208     int len = str.length();
    209     int count = 1;
    210     for(int i=0;i<len;i++){
    211         if(i+1 < len && str[i+1] == str[i]){
    212             count++;
    213         }else{
    214             res.push_back(str[i]);
    215             res.push_back(count+'0');
    216             count = 1;
    217         }
    218     }
    219     return res;
    220 }
    221 
    222 }
    223 ;
    224 
    225 #endif /* MYSTRING_H_ */
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  • 原文地址:https://www.cnblogs.com/wxquare/p/6531303.html
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