AstringSof lowercase letters is given. We want to partition thisstring into as many parts as possible so that each letter appears in at mostone part, and return a list of integers representing the size of these parts.

Example1:

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 mostone part.

A partition like "ababcbacadefegde","hijhklij" is incorrect, because it splits S into less parts.

Note:

1.      Swill have length in range[1, 500].

2.      Swill consist of lowercase letters ('a'to'z') only.

從這一題開始有些難度了，這題要求你把一個字串劃分為儘可能多的片段，每個片段儘可能包含多的字母

這一題我是使用貪心演算法的思想解決的，對於字串S而言，其最左邊的一組一定包含S最左邊的第一個字母，然後檢索該字母在S中最右邊的位置，這裡假設為i，如果從S最左邊到i之間有一個字母最右邊的索引在i右邊，那麼更新i，繼續遍歷，直到S最左端到i的字母已經全部遍歷了一遍，這個時候0~i就是一個最小的分組，將S切分，對i+1到S最右端繼續以上操作

執行該操作直到S切分完畢

class Solution {
public:
  int divide(string S, int start, int end,int letterRight[26])
{
	int left, right;
	left = start;
	right = letterRight[S[left] - 'a'];
	for (int i = left+1; i <= right; i++)
	{
		if (letterRight[S[i]-'a'] > right) right = letterRight[S[i] - 'a'];
	}
	return right - left+1;

}

vector<int> partitionLabels(string S) {
	int letterRight[26];
	vector<int> result;

	for (int i = 0; i < 26; i++)
	{
		int t = S.find_last_of((char)( i + 'a'));
		letterRight[i] = t;
	}
	int start = 0; int end = S.size(); int tmp;
	while (start < end)
	{
		tmp = divide(S, start, end, letterRight);
		result.push_back(tmp);
		start += tmp;
	}
	return result;
}
};