【貪心】B028_劃分字母區間（記錄結束位置）

一、題目描述

A string S 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.

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.

二、題解

方法一：枚舉區間

• 對於每一個字符，都會有一個終止位置，我們用 int end[] 數組每個字符記錄最後出現的位置。
• 枚舉區間 [0, end[s[l]-'a']，在區間內可能會有字符的終止位置比當前字符還要遠，所以我們需要更新每一次查找的結束位置。

public List<Integer> partitionLabels(String S) {
    char[] s = S.toCharArray();
    int[] end = new int[26];
    int N = s.length;
    for (int i = 0; i < N; i++) {
        end[s[i]-'a'] = i;
    }
    List<Integer> list = new LinkedList<>();
    
    for (int l = 0; l < N;) {
        int r = end[s[l]-'a'];
        for (int k = l + 1; k < r; k++) {
            if (end[s[k]-'a'] > r)
                r = end[s[k]-'a'];
        }
        list.add(r - l + 1);
        l = r + 1;
    }
    return list;
}

複雜度分析

• 時間複雜度：$O(n)$，
• 空間複雜度：$O(1)$，