<p>You are given a string <code>s</code>. We want to partition the string into as many parts as possible so that each letter appears in at most one part.</p>

<p>Note that the partition is done so that after concatenating all the parts in order, the resultant string should be <code>s</code>.</p>

<p>Return <em>a list of integers representing the size of these parts</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ababcbacadefegdehijhklij&quot;
<strong>Output:</strong> [9,7,8]
<strong>Explanation:</strong>
The partition is &quot;ababcbaca&quot;, &quot;defegde&quot;, &quot;hijhklij&quot;.
This is a partition so that each letter appears in at most one part.
A partition like &quot;ababcbacadefegde&quot;, &quot;hijhklij&quot; is incorrect, because it splits s into less parts.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;eccbbbbdec&quot;
<strong>Output:</strong> [10]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    vector<int> partitionLabels(string s) {
        vector<int> Labels;
        vector<array<int, 2>> positions = findPosition(s);
        int prev = positions[0][1];
        int prevBegin = positions[0][0];
        for (int i = 0; i < s.size(); ++i) {
            if (positions[i][0] <= prev && i != s.size()-1) {
                if (positions[i][1] <= prev) {
                    continue;
                }
                else {
                    prev = positions[i][1];
                }
            } 
            else {
                Labels.push_back(prev-prevBegin+1);
                prev = positions[i][1];
                prevBegin = positions[i][0];
            }
        }
        if (prev == prevBegin) {
            Labels.push_back(1);
        }
        return Labels;
    }
private:
    vector<array<int, 2>> findPosition(const string &s) {
        unordered_map<char, array<int, 2>> charPositions;
        vector<array<int, 2>> positions(s.size());

        // 初始化每个字符的第一次和最后一次出现的位置
        for (int i = 0; i < s.size(); ++i) {
            if (charPositions.find(s[i]) == charPositions.end()) {
                charPositions[s[i]] = {i, i}; // 第一次出现
            } else {
                charPositions[s[i]][1] = i; // 更新最后一次出现的位置
            }
        }

        // 构建结果向量
        for (int i = 0; i < s.size(); ++i) {
            positions[i] = charPositions[s[i]];
        }

        return positions;
    }
};
10000