[LeetCode] Partition Labels

Posted 2020-10-21 immjc

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:

1. S will have length in range [1, 500].
2. S will consist of lowercase letters (‘a‘ to ‘z‘) only.

尽可能多的找出分区，要求每个分区中的每个字母出现次数最多。

通过Example可以大致了解题目含义。

贪心算法思路：

1、使用一个map来存储每个字母出现的最大索引值。

如例子所示：

a-8, b-5, c-7, d-14, e-15, f-11, g-13, h-19, i-22, j-23, k-20, l-21

2、遍历数组，要求在当前字母所包含的分区最大。

如ababcbaca.

由map可知，a-8，在索引0-8之间所有字母在map中的最大索引不超过8。所以可以构成一个分区

这时从索引9开始遍历。d-14，在所有9-14之间的字母中，存在e的最大索引超过了14，所以令分区扩大为9-15。

剩下的部分按照这个思路遍历即可。

3、将各分区的大小放去res数组中即可。

class Solution {
public:
    vector<int> partitionLabels(string S) {
        vector<int> res;
        unordered_map<char, int> max_pos;
        for (int i = 0; i < S.size(); i++) {
            max_pos[S[i]] = i;
        }
        for (int i = 0; i < S.size();) {
            int beg = i;
            int end = max_pos[S[i]];
            for (int j = i + 1; j < end; j++) {
                end = max(end, max_pos[S[j]]);
            }
            res.push_back(end - beg + 1);
            i = end + 1;
        }
        return res;
    }
};
// 10 ms