Remove given characters in input string, the relative order of other characters should be remained. Return the new string after deletion.
Assumptions
The given input string is not null.
The characters to be removed is given by another string, it is guranteed to be not null.
Examples
input = "abcd", t = "ab", delete all instances of 'a' and 'b', the result is "cd".
time: o(n) space o(n)
public String remove(String input, String t) {
// Write your solution here
char[] inputs = input.toCharArray() ;
char[] ts = t.toCharArray() ;
//put all the ts input hashset
Set<Character> set = new HashSet<>(ts.length) ;
StringBuilder sb = new StringBuilder() ;
for (int i = 0; i < ts.length; i++) {
set.add(ts[i]);
}
for (int i = 0; i <inputs.length ; i++) {
if (!set.contains(inputs[i])){
sb.append(inputs[i]);
}
}
return sb.toString();
}
//another method using two pointers: valid area: [0, slow)
public String remove_twopointer(String input, String t) {
// Write your solution here
int slow = 0, fast = 0 ;
//put the t into hashset
Set<Character> set = new HashSet<>(t.length());
char[] inputs = input.toCharArray() ;
for (int k = 0; k < t.length(); k++) {
set.add(t.charAt(k));
}
//only need to check the fast
/* d
* s t u d
* i i i-i
* j j j j j
String
public String(char[] value,
int offset,
int count)
Allocates a new String that contains characters from a subarray of the character array argument.
The offset argument is the index of the first character of the subarray and the count argument specifies the length of the subarray.
The contents of the subarray are copied;
subsequent modification of the character array does not affect the newly created string.
Parameters:
value - Array that is the source of characters
offset - The initial offset
count - The length
* */
while (fast<input.length()){
if (set.contains(input.charAt(fast))){
fast++ ;
} else{
inputs[slow++] = inputs[fast++] ;
}
}
return new String(inputs, 0, slow);
}