0763. 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;
    }
}