LeetCode-Partition Labels

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Description：
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.

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 letters (‘a’ to ‘z’) only.

题意：将一个字符串进行最大数量的划分，要求划分后每个相同字符只出现在最多一个划分中；返回划分后的各部分的长度；

解法：遍历字符串 $S$，找到字符 $S_i$最后出现的位置 $ed$，之后我们在 $[i,ed]$范围内找出所有字符最后出现位置的最大值 $maxIndex$，这个时候有下面两种情况；

• 如果 $maxIndex == ed$，则范围 $[i, ed]$就是其中一个划分；
• 否则，从位置 $ed + 1$开始，直到 $ed == maxIndex$，其中 $maxIndex$是遍历过程中每个字符最后出现位置的最大值；

Java

class Solution {
    public List<Integer> partitionLabels(String S) {
        int[] map = new int[26];
        for (int i = 0; i < S.length(); i++) {
            map[S.charAt(i) - 'a'] = i;
        } //计算每个字符最后出现的位置
        int st = 0;
        int ed = 0;
        List<Integer> res = new ArrayList<>();
        while (st < S.length()) {
            ed = map[S.charAt(st) - 'a'];
            int maxIndex = st;
            for (int i = st; i <= ed; i++) {
                if (maxIndex < map[S.charAt(i) - 'a']) {
                    maxIndex = map[S.charAt(i) - 'a'];
                }
            } //计算范围[st, ed]内字符最后出现位置的最大值
            if (maxIndex > ed) {
                while (maxIndex != ed && ed != S.length()) {
                    ed++;
                    if (maxIndex < map[S.charAt(ed) - 'a']) {
                        maxIndex = map[S.charAt(ed) - 'a'];
                    }
                }
            } 
            res.add(ed - st + 1);
            st = ed + 1;
        }
        
        return res;
    }
}