数据结构中的字符串：概念、操作与实战
第一部分 字符串的分类及常见形式
字符串是由零个或多个字符组成的有限序列，是编程中最基础也最重要的数据结构之一。
1. C语言中的字符串表示
字符数组形式
char str1[10] = {'H', 'e', 'l', 'l', 'o', '\0'};
字符串字面量
char str2[] = "Hello World";
动态分配字符串
char *str3 = (char*)malloc(20 * sizeof(char));
strcpy(str3, "Dynamic String");
2. 字符串的存储方式
固定长度存储
#define MAX_LEN 100
char fixedStr[MAX_LEN];
动态长度存储
typedef struct {
char *data;
int length;
} DynamicString;
3. 特殊字符串类型
宽字符字符串
#include <wchar.h>
wchar_t wideStr[] = L"宽字符字符串";
多字节字符串
char mbStr[] = "多字节字符串";
第二部分 字符串常见操作
1. 基本操作函数
字符串长度
size_t my_strlen(const char *str) {
size_t len = 0;
while (*str++) len++;
return len;
}
字符串复制
char* my_strcpy(char *dest, const char *src) {
char *ret = dest;
while ((*dest++ = *src++));
return ret;
}
字符串连接
char* my_strcat(char *dest, const char *src) {
char *ret = dest;
while (*dest) dest++;
while ((*dest++ = *src++));
return ret;
}
字符串比较
int my_strcmp(const char *s1, const char *s2) {
while (*s1 && (*s1 == *s2)) {
s1++;
s2++;
}
return *(unsigned char*)s1 - *(unsigned char*)s2;
}
2. 字符串查找与分割
查找字符
char* my_strchr(const char *str, int c) {
while (*str != (char)c) {
if (!*str++) return NULL;
}
return (char*)str;
}
查找子串
char* my_strstr(const char *haystack, const char *needle) {
if (!*needle) return (char*)haystack;
for (; *haystack; haystack++) {
if (*haystack == *needle) {
const char *h = haystack, *n = needle;
while (*h && *n && *h == *n) {
h++;
n++;
}
if (!*n) return (char*)haystack;
}
}
return NULL;
}
字符串分割
char* my_strtok(char *str, const char *delim) {
static char *last = NULL;
if (str) last = str;
if (!last || !*last) return NULL;
char *start = last;
while (*last && !strchr(delim, *last)) last++;
if (*last) {
*last = '\0';
last++;
} else {
last = NULL;
}
return start;
}
3. 字符串转换与格式化
字符串转数字
int my_atoi(const char *str) {
int sign = 1, result = 0;
while (*str == ' ') str++;
if (*str == '-' || *str == '+') {
sign = (*str++ == '-') ? -1 : 1;
}
while (*str >= '0' && *str <= '9') {
result = result * 10 + (*str - '0');
str++;
}
return sign * result;
}
数字转字符串
void my_itoa(int num, char *str) {
int i = 0, sign = num;
if (sign < 0) num = -num;
do {
str[i++] = num % 10 + '0';
} while ((num /= 10) > 0);
if (sign < 0) str[i++] = '-';
str[i] = '\0';
// 反转字符串
int len = i;
for (int j = 0; j < len/2; j++) {
char temp = str[j];
str[j] = str[len-j-1];
str[len-j-1] = temp;
}
}
字符串格式化
int my_sprintf(char *str, const char *format, ...) {
va_list args;
va_start(args, format);
int len = vsprintf(str, format, args);
va_end(args);
return len;
}
第三部分 字符串编程题例子
1. 反转字符串
void reverseString(char* s, int sSize) {
int left = 0, right = sSize - 1;
while (left < right) {
char temp = s[left];
s[left++] = s[right];
s[right--] = temp;
}
}
2. 验证回文字符串
bool isPalindrome(char* s) {
int left = 0, right = strlen(s) - 1;
while (left < right) {
while (left < right && !isalnum(s[left])) left++;
while (left < right && !isalnum(s[right])) right--;
if (tolower(s[left++]) != tolower(s[right--])) {
return false;
}
}
return true;
}
3. 字符串转换整数 (atoi)
int myAtoi(char* str) {
long result = 0;
int sign = 1;
while (*str == ' ') str++;
if (*str == '-' || *str == '+') {
sign = (*str++ == '-') ? -1 : 1;
}
while (*str >= '0' && *str <= '9') {
result = result * 10 + (*str - '0');
if (sign == 1 && result > INT_MAX) return INT_MAX;
if (sign == -1 && -result < INT_MIN) return INT_MIN;
str++;
}
return (int)(sign * result);
}
4. 最长公共前缀
char* longestCommonPrefix(char** strs, int strsSize) {
if (strsSize == 0) return "";
for (int i = 0; ; i++) {
char c = strs[0][i];
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != c || strs[j][i] == '\0') {
char *result = malloc(i + 1);
strncpy(result, strs[0], i);
result[i] = '\0';
return result;
}
}
if (c == '\0') break;
}
return strdup(strs[0]);
}
5. 字符串的排列组合
bool checkInclusion(char* s1, char* s2) {
int len1 = strlen(s1), len2 = strlen(s2);
if (len1 > len2) return false;
int count1[26] = {0}, count2[26] = {0};
for (int i = 0; i < len1; i++) {
count1[s1[i] - 'a']++;
count2[s2[i] - 'a']++;
}
int matches = 0;
for (int i = 0; i < 26; i++) {
if (count1[i] == count2[i]) matches++;
}
for (int i = len1; i < len2; i++) {
if (matches == 26) return true;
int left = s2[i - len1] - 'a';
count2[left]--;
if (count2[left] == count1[left]) {
matches++;
} else if (count2[left] == count1[left] - 1) {
matches--;
}
int right = s2[i] - 'a';
count2[right]++;
if (count2[right] == count1[right]) {
matches++;
} else if (count2[right] == count1[right] + 1) {
matches--;
}
}
return matches == 26;
}
6. 字符串相乘
char* multiply(char* num1, char* num2) {
int len1 = strlen(num1), len2 = strlen(num2);
int len = len1 + len2;
int *result = calloc(len, sizeof(int));
char *final = malloc(len + 1);
for (int i = len1 - 1; i >= 0; i--) {
for (int j = len2 - 1; j >= 0; j--) {
int product = (num1[i] - '0') * (num2[j] - '0');
int sum = product + result[i + j + 1];
result[i + j] += sum / 10;
result[i + j + 1] = sum % 10;
}
}
int index = 0;
while (index < len && result[index] == 0) index++;
if (index == len) {
final[0] = '0';
final[1] = '\0';
return final;
}
int pos = 0;
while (index < len) {
final[pos++] = result[index++] + '0';
}
final[pos] = '\0';
free(result);
return final;
}
7. 正则表达式匹配
bool isMatch(char* s, char* p) {
if (*p == '\0') return *s == '\0';
bool first_match = (*s != '\0') && (*p == *s || *p == '.');
if (*(p + 1) == '*') {
return isMatch(s, p + 2) || (first_match && isMatch(s + 1, p));
} else {
return first_match && isMatch(s + 1, p + 1);
}
}
字符串处理是编程中最常见的任务之一，几乎所有的应用程序都会涉及到字符串操作。掌握字符串的基本操作和常见算法，对于解决实际问题至关重要。通过练习这些典型题目，可以深入理解字符串的特性和处理技巧，提高编程能力和算法思维。
