leetcode WordBreak2实现算法

本文算法对于：大数据aaaaaaaaaaaaaaaaaaaaaaa;无法通过
时间越界Last executed input:"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

public ArrayList<String> list = new ArrayList<String>();
public void wordBreak(String s, Set<String> dict, String tmp) {
        if (s == null || s.length() == 0)
            return;
        int length = s.length();
        int beg = 0, end = length;
        int index = 0;
        while (beg < end) {
            end = length - index;
            if (beg < end) {
                String str = s.substring(beg, end);
                if (dict.contains(str)) {
                    if (end == length) {
                        list.add((tmp + " " + str).trim());
                    } else {
                        wordBreak(s.substring(end), dict, tmp + " " + str);
                    }
                }
                index++;
            } else {
                break;
            }
        }
    }