字符串去重C语言实现

字符串去重经常会考的笔试题目，这里列出几种常用的方法

更详细的解释（C++版本）请参考http://hawstein.com/posts/1.3.html

解法一：取第一个字符然后遍历后面所有字符，若有重复的则将后面的字符设置为'\0'

//将重复字符设置为'\0'void RemoveDuplicate(char *str){int i, j, k, len;len = strlen(str);for(i = k = 0; i < len; i++){if(str[i]){str[k++] = str[i];for(j = i + 1; j < len; j++)if(str[j] == str[i])str[j] = '\0';}}str[k] = '\0';}

解法二：设置一个标记数组，检查是否有重复字符出现，若没有出现过则插入字符串

void RemoveDuplicate(char *s){char check[256] = { 0 };int i, j, len;len = strlen(s);for(i = j = 0; i < len; i++){if(check[s[i]] == 0){s[j++] = s[i];check[s[i]] = 1;}}s[j] = '\0';}

进一步优化，这里标记数组用了256个字节，我们可以用含有8个整型元素的数组来表示

void RemoveDuplicate(char *s){int i, j, len, remainder;int check[8] = {0};len = strlen(s);for(i = j = 0; i < len; i++){remainder = s[i] % 32;if((check[s[i] >> 5] & (1 << remainder)) == 0){s[j++] = s[i];check[s[i] >> 5] |= (1 << remainder);}}s[j] = '\0';}

继续压缩问题，如果字符串中只出现a~z之间的小写字母，可用一个整型变量表示

void RemoveDuplicate(char *s){int i, j, val, check;j = check = 0;for(i = 0; s[i]; i++){val = s[i] - 'a';if((check & (1 << val)) == 0){s[j++] = s[i];check |= 1 << val;}}s[j] = '\0';}