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_316.java
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package com.fishercoder.solutions;
import java.util.*;
/**
* 316. Remove Duplicate Letters
*
* 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"
*/
public class _316 {
/**credit: https://discuss.leetcode.com/topic/32259/java-solution-using-stack-with-comments/2*/
public String removeDuplicateLetters_use_stack(String s) {
int[] res = new int[26]; //will contain number of occurences of character (i+'a')
boolean[] visited = new boolean[26]; //will contain if character (i+'a') is present in current result Stack
char[] ch = s.toCharArray();
for(char c: ch){ //count number of occurences of character
res[c-'a']++;
}
Deque<Character> st = new ArrayDeque<>(); // answer stack
int index;
for(char c : ch){
index= c-'a';
res[index]--; //decrement number of characters remaining in the string to be analysed
if(visited[index]) {//if character is already present in stack, dont bother
continue;
}
//if current character is smaller than last character in stack which occurs later in the string again
//it can be removed and added later e.g stack = bc remaining string abc then a can pop b and then c
while(!st.isEmpty() && c<st.peek() && res[st.peek()-'a']!=0){
visited[st.pop()-'a']=false;
}
st.push(c); //add current character and mark it as visited
visited[index]=true;
}
StringBuilder sb = new StringBuilder();
//pop character from stack and build answer string from back
while(!st.isEmpty()){
sb.insert(0,st.pop());
}
return sb.toString();
}
/**Credit: https://discuss.leetcode.com/topic/31404/a-short-o-n-recursive-greedy-solution*/
public String removeDuplicateLetters(String s) {
int[] count = new int[26];
int pos = 0; // the position for the smallest s[i]
for (int i = 0; i < s.length(); i++) count[s.charAt(i) - 'a']++;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) < s.charAt(pos)) pos = i;
count[s.charAt(i) - 'a']--;
if (count[s.charAt(i) - 'a'] == 0) break;
}
return s.length() == 0 ? "" : s.charAt(pos) + removeDuplicateLetters(s.substring(pos + 1).replaceAll("" + s.charAt(pos), ""));
}
}