Description
You are given a string s consisting of lowercase English letters. A duplicate removal consists of choosing two adjacent and equal letters and removing them.
We repeatedly make duplicate removals on s until we no longer can.
Return the final string after all such duplicate removals have been made. It can be proven that the answer is unique.
Example 1:
Input: s = "abbaca"
Output: "ca"
Explanation:
For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".
Example 2:
Input: s = "azxxzy"
Output: "ay"
Constraints:
1 <= s.length <= 105s consists of lowercase English letters.
Solutions
Solution 1
PythonJavaC++GoRustJavaScriptC
class Solution:
def removeDuplicates(self, s: str) -> str:
stk = []
for c in s:
if stk and stk[-1] == c:
stk.pop()
else:
stk.append(c)
return ''.join(stk)(code-box)
class Solution {
public String removeDuplicates(String s) {
StringBuilder sb = new StringBuilder();
for (char c : s.toCharArray()) {
if (sb.length() > 0 && sb.charAt(sb.length() - 1) == c) {
sb.deleteCharAt(sb.length() - 1);
} else {
sb.append(c);
}
}
return sb.toString();
}
}(code-box)
class Solution {
public:
string removeDuplicates(string s) {
string stk;
for (char c : s) {
if (!stk.empty() && stk[stk.size() - 1] == c) {
stk.pop_back();
} else {
stk += c;
}
}
return stk;
}
};(code-box)
func removeDuplicates(s string) string {
stk := []rune{}
for _, c := range s {
if len(stk) > 0 && stk[len(stk)-1] == c {
stk = stk[:len(stk)-1]
} else {
stk = append(stk, c)
}
}
return string(stk)
}(code-box)
impl Solution {
pub fn remove_duplicates(s: String) -> String {
let mut stack = Vec::new();
for c in s.chars() {
if !stack.is_empty() && *stack.last().unwrap() == c {
stack.pop();
} else {
stack.push(c);
}
}
stack.into_iter().collect()
}
}(code-box)
/**
* @param {string} s
* @return {string}
*/
var removeDuplicates = function (s) {
const stk = [];
for (const c of s) {
if (stk.length && stk[stk.length - 1] == c) {
stk.pop();
} else {
stk.push(c);
}
}
return stk.join('');
};(code-box)
char* removeDuplicates(char* s) {
int n = strlen(s);
char* stack = malloc(sizeof(char) * (n + 1));
int i = 0;
for (int j = 0; j < n; j++) {
char c = s[j];
if (i && stack[i - 1] == c) {
i--;
} else {
stack[i++] = c;
}
}
stack[i] = '\0';
return stack;
}(code-box)
