关于LeetCode中Word Pattern一题的理解

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题目如下：

Given a pattern and a stringstr, find ifstr follows the same pattern.

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

Examples:

pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
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.

这道题和之前的Isomorphic String一题可以使用相同的解决方法，我这次还是使用和上次相同的解决方案，直接把代码贴出来了，Isomorphic String的博客地址在这里：http://blog.csdn.net/zsy112371/article/details/52461066 。

 public boolean wordPattern(String pattern, String str) {        String[] origin = str.split(" ");        if(pattern.length() != origin.length){            return false;        }else{            int mod_length = pattern.length();            int[] mod = new int[mod_length];            int[] mod2 = new int[mod_length];                        HashMap<String,Integer> comparedMap = new HashMap<String,Integer>();            HashMap<String,Integer> comparedMap2 = new HashMap<String,Integer>();                        for(int i=0;i<mod_length;i++){                String pattern_char = pattern.charAt(i)+"";                String origin_char = origin[i];                                if(!comparedMap.containsKey(pattern_char)){                    comparedMap.put(pattern_char,i);                    mod[i] = i;                }else{                    mod[i] = comparedMap.get(pattern_char).intValue();                }                                if(!comparedMap2.containsKey(origin_char)){                    comparedMap2.put(origin_char,i);                    mod2[i] = i;                }else{                    mod2[i] = comparedMap2.get(origin_char).intValue();                }                                if(mod[i]!=mod2[i]){                    return false;                }            }                        return true;        }    }

然后再来看一下讨论区的回答，首先是下面的解决方案，非常简练，这个代码的解释在这里：https://discuss.leetcode.com/topic/26339/8-lines-simple-java

public boolean wordPattern(String pattern, String str) {        String[] words = str.split(" ");        if (words.length != pattern.length())            return false;        Map index = new HashMap();        for (Integer i=0; i<words.length; ++i)            if (index.put(pattern.charAt(i), i) != index.put(words[i], i))                return false;        return true;    }

然后是一种常规使用HashMap的方法，还是挺有意思的。首先，如果map中包含key值，我们就判断当前输入的pattern的第i位字符在map中对应的value值和arr[i]相不相等，如果不相等就说明不对应，应当返回false；如果map中未包含该key值，我们就判断map中是否包含值为arr[i]的value值，如果包含说明已经存在一个key与之对应，但是这个key和现在输入的pattern的第i个字符代表的key不是同一个key，说明同一个value值对应了两个不同的key值，说明不对应，返回false。代码如下所示：

 public boolean wordPattern(String pattern, String str) {        String[] arr= str.split(" ");        HashMap<Character, String> map = new HashMap<Character, String>();        if(arr.length!= pattern.length())            return false;        for(int i=0; i<arr.length; i++){            char c = pattern.charAt(i);            if(map.containsKey(c)){                if(!map.get(c).equals(arr[i]))                    return false;            }else{                if(map.containsValue(arr[i]))                    return false;                map.put(c, arr[i]);            }            }        return true;    }

基本就是这些方法了，今天感觉诸事不顺，感觉身体被掏空，托福复习的像坨shit，FXXK 。

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