Partition Labels (M)
题目
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.
题意
将给定字符串划分成尽可能多的子串,使得同一个字母只出现在一个子串中。
思路
主要思想是记录每个字母出现的最后一个位置,得到每个字母的出现区间。对于一个字母的出现区间S,如果其中有字母的结束位置大于S的右端点,那么需要更新右端点,直到右端点对应字母的结束位置就是右端点本身,这样就找到了一个符合条件的子串。
代码实现
Java
class Solution {
public List<Integer> partitionLabels(String S) {
List<Integer> ans = new ArrayList<>();
int[] ends = new int[26];
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
ends[c - 'a'] = Math.max(ends[c - 'a'], i);
}
int start = 0, end = 0;
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
end = Math.max(end, ends[c - 'a']);
if (end == i) {
ans.add(end - start + 1);
start = i + 1;
}
}
return ans;
}
}