LeetCode -- Word Pattern

题目描述：


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:
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.


给一个模式字符串：pattern，和一个包含了很多单词由空格隔开组成的字符串str，判断是否符合同样的结构。
符合同样的结构即每个单词的对应是不变的。


思路：
1. 如果长度不一样肯定不同构。
2. 设pattern[i]为pattern中的第i个字符，arr为str分隔后的单词数组。
使用两个字典dict1和dict2，遍历pattern过程中，如果pattern[i]在dict1中存在，那么arr[i]在dict2中也一定存在，并且dict1[pattern[i]]与arr[i]是相等的；而如果pattern[i]在dict1中不存在，那么arr[i]在dict2中也一定不应该存在。




实现代码：

public class Solution {    public bool WordPattern(string pattern, string str) {        var i = 0;    var arr = str.Split(' ').Where(x=>!string.IsNullOrWhiteSpace(x)).ToList();    var dict1 = new Dictionary<char, string>();    var dict2 = new Dictionary<string, char>();        if(pattern.Length != arr.Count){    return false;    }        while(i < pattern.Length)    {    if(!dict1.ContainsKey(pattern[i])){    if(dict2.ContainsKey(arr[i])){    return false;    }        dict1.Add(pattern[i] , arr[i]);    dict2.Add(arr[i], pattern[i]);    }    else{    if(!dict2.ContainsKey(arr[i])){    return false;    }    if(arr[i] != dict1[pattern[i]]){    return false;    }    }        i++;    }        return true;    }}