Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.
Example:
Given "bcabc"
Return "abc"
Given "cbacdcbc"
Return "acdb"
The idea is to keep a Monotonous stack
when a letter comes, if it is already in the stack, continue;
if it is not in the stack, we pop out the top element of the stack util the top element is smaller than the letter. (if a letter is in its last occurrence, then we need to break because each letter should exists at least once).
code:
public class Solution {
public String removeDuplicateLetters(String s) {
int[] last = new int[26];
for(int i = 0 ; i < s.length(); i++){
char c = s.charAt(i);
last[c - 'a'] = i;
}
LinkedList<Character> list = new LinkedList<>();
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if(list.contains(c)) continue;
while(list.size() > 0 && list.getLast() >= c && last[list.getLast() - 'a'] > i){
list.removeLast();
}
if(!list.contains(c)) list.add(c);
}
StringBuilder ret = new StringBuilder();
for(char c : list){
ret.append(c);
}
return ret.toString();
}
}
本文介绍了一种算法,该算法可以去除给定字符串中的重复字符,并确保处理后的字符串保持最小的字典序。通过使用单调栈的概念,文章详细解释了如何在遍历字符串的同时判断何时弹出栈顶元素,以确保最终结果既不含重复字符又保持最小字典序。
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