[Leetcode] 763. Partition Labels 解题报告

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本文介绍了一种高效的算法，用于将一个由小写字母组成的字符串尽可能多地分割成若干部分，确保每个字母只出现在一个部分中。通过两次扫描字符串并利用哈希表记录字符最后一次出现的位置来实现这一目标。

题目：

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.

思路：

我们定义一个哈希表，计算每个字符在该字符串中最后出现的位置，然后再次顺序扫描字符串，并且更新已经扫描过的字符的最后出现位置。一旦发现当前扫描位置与目前的最终出现位置相等，就说明可以分割了，此时返回该段子串的长度即可。算法的时间复杂度是O(n)，空间复杂度也是O(n)。

代码：

class Solution {
public:
    vector<int> partitionLabels(string S) {
        unordered_map<char, int> hash;  // map from char to its last index
        for (int i = 0; i < S.length(); ++i) {
            hash[S[i]] = i;
        }
        int last_end = -1, max_end = -1;
        vector<int> ret;
        for (int start = 0; start < S.length(); ++start) {
            max_end = max(max_end, hash[S[start]]);
            if (max_end == start) {
                int length = max_end - last_end;
                ret.push_back(length);
                last_end = start;
            }
        }
        return ret;
    }
};