763 Partition Labels
763. Partition Labels
1. Question
A stringS
of lowercase 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 letters ('a'
to'z'
) only.
2. Implementation
(1) Two Pointers
class Solution {
public List<Integer> partitionLabels(String S) {
List<Integer> res = new ArrayList<>();
if (S == null || S.length() == 0) {
return res;
}
// map记录s中的character最后出现的index
int[] map = new int[26];
for (int i = 0; i < S.length(); i++) {
map[S.charAt(i) - 'a'] = i;
}
int last = 0, start = 0;
for (int i = 0; i < S.length(); i++) {
last = Math.max(last, map[S.charAt(i) - 'a']);
if (last == i) {
res.add(last - start + 1);
start = i + 1;
}
}
return res;
}
}
3. Time & Space Complexity
Two Pointers: 时间复杂度O(n), 空间复杂度O(n)
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