从字符串中删除重复的字符

设计一个算法去移除字符串中重复的字符，但是只可以使用一个或两个变量的额外空间。

Design an algorithm and write code to remove the duplicate characters in a

string without using any additional buffer. NOTE: One or two additional variables

is fine. An extra copy of the array is not. pg 22

为算法写出测试用例。

FOLLOW UP

Write the test cases for this method.

解：

SOLUTION

First, ask yourself, what does the interviewer mean by an additional buffer? Can we use an

additional array of constant size?

不用额外空间的解法：

将原数组分为两边，一边为不重复的字符串a，另一边为要处理的字符串b。

对于b中的字符，检查他是否在a中出现，如果出现了，处理b中的下一个字符，如果没有出现，将其添加到a中。

时间复杂度为O(N2).

初始时a中有一个字符。

Algorithm—No (Large) Additional Memory:

1. For each character, check if it is duplicate of already found characters.

2. Skip duplicate characters and update the non duplicate characters.

Time complexity is O(N2).

void removeDuplicates(char *str) {
if (!str)
return;
int len = strlen(str);
if (len < 2)
return;
int tail = 1;
for (int i = 1; i < len; ++i) {
int j;
for (j = 0; j < tail; ++j)
if (str[i] == str[j])
break;
if (j == tail) {
str[tail] = str[i];
++tail;
}
}
str[tail] = 0;
}

Test Cases:

1. String does not contain any duplicates, e.g.: abcd

2. String contains all duplicates, e.g.: aaaa

3. Null string

4. String with all continuous duplicates, e.g.: aaabbb

5. String with non-contiguous duplicate, e.g.: abababa

使用额外空间的算法：

使用hit数组记录字符是否重复，如果为true表示在未重复的字符串中。

Algorithm—With Additional Memory of Constant Size

void removeDuplicatesEff(char *str) {
if (!str)
return;
int len = strlen(str);
if (len < 2)
return;
bool hit[256];
for(int i = 0; i < 256; ++i)
hit[i] = false;
hit[str[0]] = true;
int tail = 1;
for (int i = 1; i < len; ++i) {
if (!hit[str[i]]) {
str[tail] = str[i];
++tail;
hit[str[i]] = true;
}
}
str[tail] = 0;
}

Test Cases:

1. String does not contain any duplicates, e.g.: abcd

2. String contains all duplicates, e.g.: aaaa

3. Null string

4. Empty string

5. String with all continuous duplicates, e.g.: aaabbb

6. String with non-contiguous duplicate, e.g.: abababa