leetcode講解--763. Partition Labels

題目

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:

1. S will have length in range [1, 500].
2. S will consist of lowercase letters ('a' to 'z') only.

題目地址

講解

這道題我竟然一遍過，之前作題或多或少有些小錯誤。這道題我一上手就發現應該儘可能阻止對數據的屢次遍歷，因此若是能遍歷一遍獲得一些有用的信息就行了。我用的解法是，創建一個map存儲每一個字母的首尾位置（固然實際操做中我用的是兩個map分別存儲字母第一次出現的位置和最後一次出現的位置）。而後若是這一段之間出現了一個字母，它的尾部位置比框住它的這個字母更大，就更新尾部位置，直到尾部位置沒法再擴大爲止。

Java代碼

class Solution {
    public List<Integer> partitionLabels(String S) {
        List<Integer> result = new ArrayList<>();
        char[] c = S.toCharArray();
        Map<Character, Integer> mapStart = new HashMap<>();
        Map<Character, Integer> mapEnd = new HashMap<>();
        for(int i=0;i<c.length;i++){
            if(mapStart.get(c[i])==null){
                mapStart.put(c[i], i);
                mapEnd.put(c[i], i);
            }else{
                mapEnd.put(c[i], i);
            }
        }
        int beginIndex=0;
        while(beginIndex<c.length){
            int endIndex=mapEnd.get(c[beginIndex]);
            for(int i=beginIndex;i<endIndex;i++){
                if(mapEnd.get(c[i])>endIndex){
                    endIndex = mapEnd.get(c[i]);
                }
            }
            result.add(endIndex-beginIndex+1);
            beginIndex = endIndex+1;
        }
        return result;
    }
}