A string S of lowercase English letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
Example 1:
Input: S = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
Note:
S will have length in range [1, 500].
S will consist of lowercase English letters ('a' to 'z') only.
Solution:
brute force
class Solution {
public List<Integer> partitionLabels(String S) {
List<Set<Character>> list = new ArrayList();
List<Integer> result = new ArrayList();
for (char c : S.toCharArray()) {
boolean duplicate = false;
for (int i = 0; i < list.size(); i ++) {
Set<Character> set = list.get(i);
if (set.contains(c)) {
merge(list, result, i);
list = list.subList(0, i + 1);
result = result.subList(0, i + 1);
duplicate = true;
break;
}
}
if (!duplicate) {
list.add(new HashSet());
result.add(0);
}
list.get(list.size() - 1).add(c);
result.set(result.size() - 1, result.get(result.size() - 1) + 1);
}
return result;
}
private void merge(List<Set<Character>> list, List<Integer> result, int i) {
for (int j = i + 1; j < list.size(); j ++) {
list.get(i).addAll(list.get(j));
result.set(i, result.get(j) + result.get(i));
}
}
}
Solution 2:
Two pointer:
public List<Integer> partitionLabels(String S) {
Integer[] positions = new Integer[26];
char[] chs = S.toCharArray ();
for (int i = 0; i < chs.length; i++)
positions[chs[i] - 'a'] = i;
List<Integer> resLs = new ArrayList<> ();
int pos = 0, end = 0, anchor = 0;
while (pos < chs.length) {
end = Math.max (positions[chs[pos] - 'a'], end);
if (pos == end) {
resLs.add (pos - anchor + 1);
anchor = pos + 1;
}
pos++;
}
return resLs;
}