字符串去重C语言实现
http://blog.csdn.net/zitong_ccnu/article/details/11820511
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';
- }
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