题目地址: https://leetcode.com/problems/partition-labels/
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.字符串 S 由小写字母组成。我们要把这个字符串划分为尽可能多的片段，同一个字母只会出现在其中的一个片段。返回一个表示每个字符串片段的长度的列表。
这道题的意思是：同一个字母只会出现在其中的一个片段，但是这个字符串划分为尽可能多的片段
所以，只需要把找到第一个字母’a’的最后出现的位置 x, 遍历从到 [0, x] 的所以字母的最后出现的位置为 end，这个片段就是 [0, end]
1、记录所有字母的最后一个位置
2、不断循环更新该片段的所有字母最后出现的位置
class Solution { public List<Integer> partitionLabels(String S) { List<Integer> res = new ArrayList<>(); if (S == null || S.length() <= 0) return res; int[] index = new int[128]; for (int i = 0; i < S.length(); i++) { index[S.charAt(i)] = i; } int start = 0; while (start < S.length()) { int pre = start; start = pH(S, index, pre) + 1; res.add(start - pre); } return res; } public int pH(String S, int[] index, int start) { int end = index[S.charAt(start)]; for (int i = start; i <= end; i++) { if (index[S.charAt(i)] > end) end = index[S.charAt(i)]; } return end; } }执行耗时:2 ms,击败了100.00% 的Java用户 内存消耗:37.1 MB,击败了92.59% 的Java用户
