Astring S of 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.      S will have length in range [1, 500].

2.      S will consist of lowercase letters (‘a’ to ‘z’) only.

Discuss

从这一题开始有些难度了，这题要求你把一个字符串划分为尽可能多的片段，每个片段尽可能包含多的字母

这一题我是使用贪心算法的思想解决的，对于字符串S而言，其最左边的一组一定包含S最左边的第一个字母，然后检索该字母在S中最右边的位置，这里假设为i，如果从S最左边到i之间有一个字母最右边的索引在i右边，那么更新i，继续遍历，直到S最左端到i的字母已经全部遍历了一遍，这个时候0~i就是一个最小的分组，将S切分，对i+1到S最右端继续以上操作

执行该操作直到S切分完毕

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;
}
};