leetcode 763. Partition Labels

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本文介绍了一种高效的字符串分区算法，该算法旨在将给定的字符串尽可能多地拆分成多个部分，确保每个字母只在一个部分中出现。通过两次遍历实现目标，第一次遍历记录每个字符最后一次出现的位置，第二次遍历则根据这些位置信息进行分区。

763. Partition Labels

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.

题目：将字符串尽量多地拆分为几块，同一个单词只能出现在一块中。问怎么分？

解法：

先遍历S一次，知道每个单词的最后出现的位置。

然后再次遍历，每次遍历到遍历过的单词的最后出现的位置max_limit。这个max_limit则是可能会增加的。

class Solution {
public:
    vector<int> partitionLabels(string S)
    {
        vector<int> ret;
        int S_size = S.size();
        map<char, int> mp;
        for (int i = 0; i < S_size; i++)
            mp[S[i]] = i;
        
        int start_pos = 0, k = 0, max_limit = 0;
        while (k < S_size)
        {
            max_limit = mp[S[k]];
            while (k < max_limit)
            {
                max_limit = max(max_limit, mp[S[k]]);
                k ++;
            }
            ret.push_back(max_limit - start_pos + 1);
            k ++;
            start_pos = k;
        }
        return ret;
    }
};