763. Partition Labels#
1. 題目描述
題目連結
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.
2. 題目分析
這題意比較難了解,剛開始讀了好幾遍,都沒看懂。意思就是說,将一個長字元串S,進行分割多個子串,分割時必須保證每個字元隻能出現在一個子串中,并且子串必須是最小子串,就是不能繼續分割了。如"ababcbacadefegde",就還能分割: “ababcbaca”, “defegde”,是以"ababcbacadefegde"不是最小子串。
開始時,這裡有個讓我疑惑的地方是,像這種"ababcbacadefegdehijhklijxwz",前面部分分割成"ababcbaca", “defegde”, “hijhklij”,但"xwz"是作為一個整體呢,還是繼續分割成"x",“w”,"z"呢?後面經過測試,是後者。仔細想想按題目的意思,"xwz"是需要繼續分割的,“x”,“w”,"z"才是最小子串。
3. 解決思路
首先周遊整個字元串S,記錄每個字元最後出現的下标到一個maxIndex數組中。
然後,周遊字元串的時候,對于字元串S而言,從S最左邊的第一個字母,然後從數組maxIndex中找到該字母在S中最右邊出現的位置,這裡假設為end,如果從S最左邊到end之間有一個字母最右邊的索引在end右邊,那麼更新end,繼續周遊,直到S最左端到i的字母已經全部周遊了一遍,這個時候0~end就是一個最小的分組,将S切分,對end+1到S最右端繼續以上操作,重複執行該操作直到S切分完畢
4. 代碼實作(java)
public List<Integer> partitionLabels(String S) {
List<Integer> list = new LinkedList<>();
if(S == null || S.isEmpty()){
return list;
}
int len = S.length();
int[] maxIndex = new int[26];
for(int i = 0; i < len; i++) {
maxIndex[S.charAt(i) - 'a'] = i;
}
int start = 0;
int end = start;
int i;
int partitionLen = 0;
while(end < len){
i = start;
while(i <= end && end < len){
end = Math.max(end, maxIndex[S.charAt(i) - 'a']);
i++;
}
partitionLen = end - start + 1;
list.add(partitionLen);
start = end + 1;
end = start;
}
return list;
}