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：题目要求划分尽量多的区间，使得每个区间不能出现相同的字母。 那么，我们可以单独处理出每个字母(26个)在字符串中第一次出现的位置和最后一次出现的位置，然后对于这26个区间做一下合并。
class Solution {
public:
vector<int> partitionLabels(string S) {
vector< pair<int,int> > v(26);
for (int i = 0; i < 26; ++i) {
v[i].first = v[i].second = 1000;
}
for (int i = 0; i < S.size(); ++i) {
int x = S[i] - 'a';
v[x].second = i;
}
for (int i = S.size()-1; i >= 0; --i) {
int x = S[i] - 'a';
v[x].first = i;
}
sort(v.begin(), v.end());
int vis[600] = {0};
vector<int>ans;
for (int i = 0; i < 26; ++i) {
//cout << v[i].first << " " << v[i].second << endl;
if (vis[i]) continue;
int x = v[i].first;
int y = v[i].second;
if (x == 1000&& y == 1000)continue;
for (int j = i + 1; j < 26; ++j) {
int a = v[j].first;
int b = v[j].second;
if (a==1000)continue;
if (a < y) {
y = max(y, b);
vis[j]=1;
}
else break;
}
ans.push_back(y-x+1);
}
return ans;
}
};