You are given a string s
. We want to partition the string into as many parts as possible so that each letter appears in at most one part.
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.
Example 2:
Input: s = "eccbbbbdec" Output: [10]
Constraints:
1 <= s.length <= 500
-
s
consists of lowercase English letters.
Ideas: 其实跟[LeetCode] 56. Merge Intervals_Medium Tag: sort 原理一样,我们扫一遍s,得到每个char的first and last index, 然后再merge intervals, 最后返回intervals的差值加1即可。
Code:
class Solution: def partitionLabels(self, s: str) -> List[int]: d = collections.defaultdict(list) for index, c in enumerate(s): if not d[c] : d[c]= [index, index] else: d[c][-1] = index intervals = sorted(d.values()) ans = [] for s, e in intervals: if not ans or s >ans[-1][1]: ans.append([s, e]) else: ans[-1][1] = max(ans[-1][1], e) return [e - s + 1 for s, e in ans]