LeetCode.290 Word pattern

题目：

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. pattern = "abba", str = "dog dog dog dog" should return false.

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

分析：

class Solution {    public boolean wordPattern(String pattern, String str) {       //给定字符串和匹配模式，返回是否满足匹配模型        //匹配模式只有一个小些字母表示，字符串的字母由单个空格隔开组成的小写单词        HashMap<String,String> hmP=new HashMap<String,String>();        HashMap<String,String> hmS=new HashMap<String,String>();                int len=0;        for(int i=0;i<str.length();i++){            if(str.charAt(i)==' '){                len++;            }        }        if(len+1!=pattern.length())return false;                //截断字母串        String [] strs=new String[len+1];        char [] st=str.toCharArray();        int index=0;        int count=0;        for(int i=0;i<st.length;i++){            if(st[i]==' '||i==st.length-1){                if(i==st.length-1){                    //因为截取包左不包右                    i++;                }                strs[count++]=str.substring(index,i);                index=i+1;            }        }        char [] pa=pattern.toCharArray();        for(int i=0;i<pattern.length();i++){            char ch=pa[i];            String chStr=String.valueOf(ch);            //如果hash表中存在该字符则取出同字母比较            if(hmP.containsKey(chStr)){                if(!hmP.get(chStr).equals(strs[i])){                    return false;                }            }else{                //存在hmT中说明已经匹配过了                if(hmS.containsKey(strs[i])){                    return false;                }                //分别存入对方的                hmP.put(chStr,strs[i]);                hmS.put(strs[i],chStr);            }        }        return true;            }}