字符串去重經常會考的筆試題目,這裏列出幾種常用的方法
更詳細的解釋(C++版本)請參考http://hawstein.com/posts/1.3.html
解法一:取第一個字符然後遍歷後面所有字符,若有重複的則將後面的字符設置爲'\0'
//將重複字符設置爲'\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:設置一個標記數組,檢查是否有重複字符出現,若沒有出現過則插入字符串
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';
}
進一步優化,這裏標記數組用了256個字節,我們可以用含有8個整型元素的數組來表示
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';
}
繼續壓縮問題,如果字符串中只出現a~z之間的小寫字母,可用一個整型變量表示
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';
}