【LeetCode】 763. Partition Labels 划分字母区间（Medium）（JAVA）

题目地址: https://leetcode.com/problems/partition-labels/

题目描述：

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.

题目大意

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。

解题方法

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