【java】290. Word Pattern

问题原文https://leetcode-cn.com/problems/word-pattern/description/

public boolean wordPattern(String pattern, String str) {
	if(pattern == null && str == null) return true;
	if(pattern == null || str == null) return false;
		//用数字记录下模式
        int[] Ptable = new int[26];//a is 97
        StringBuilder l1 = new StringBuilder();
        for(int i = 0;i<pattern.length();i++) {
        	//这里需要注意为什么是97
        	int tmp = pattern.charAt(i)-97;
        	//注意为什么加一
        	if(Ptable[tmp] == 0) Ptable[tmp] = i+1;
        	l1.append(Ptable[tmp]);
        }
        
        int count = 1;
        StringBuilder l2 = new StringBuilder();
        Map<String,Integer> map = new HashMap<>();
        StringBuilder tmp = new StringBuilder();
        int current = 0;
        while(current<str.length()) {
        	//如果遇到空格说明上一个单词结束了，两个指针的位置也要变化了
        	if(str.charAt(current) == ' ') {
        		if(!map.containsKey(tmp.toString()))map.put(tmp.toString(), count);
        		l2.append(map.get(tmp.toString()));
        		current++;
        		count++;
        		tmp = new StringBuilder();
        	}else {
        		tmp.append(str.charAt(current++));
        	}
        } 
        //把结尾的单词加进去
        if(!map.containsKey(tmp.toString()))map.put(tmp.toString(), count);
		l2.append(map.get(tmp.toString()));
		
		System.out.println(l1);
		System.out.println(l2);
		
        if(l1.length()!=l2.length()) return false;
        for(int i = 0;i<l1.length();i++) {
        	if(l1.charAt(i) != l2.charAt(i)) return false;
        }
        return true;
    }