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.
本人思路：把字符串转化为字符串数组后，先比较长度；再用pattern里的字符替换掉字符串。全部替换后再重新转化为字符串，并与pattern字符串相比较，得出结论。
代码如下（方法略笨拙，可跳过看第二种，哈哈哈）：
public bool WordPattern(string pattern, string str) {
//char[] patternAttr=new char[pattern.Length]
char[] patternAttr=pattern.ToCharArray();
string[] strAttr=str.Split(' ');
if(patternAttr.Length!=strAttr.Length)
{
return false;
}
else
{
for(int i=;i<strAttr.Length;i++)
{
if (patternAttr[i] != ' ')
{
for(int j=i+;j<strAttr.Length;j++)
{
if(strAttr[j]==strAttr[i])
{
strAttr[j]=patternAttr[i].ToString()+"!";
patternAttr[j] = ' ';
}
}
for (int k = i + ; k < strAttr.Length;k++ )
{
if(patternAttr[k]==patternAttr[i])
{
patternAttr[i] = ' ';
}
}
strAttr[i] = patternAttr[i].ToString()+"!";
patternAttr[i] = ' ';
}
}
str=String.Join("",strAttr);
str=str.Replace("!","");
if(str==pattern)return true;
else return false;
}
}
第二种解题方法：用字典
public class Solution {
public bool WordPattern(string pattern, string str) {
string[] values = str.Split(' ');
if(pattern.Length!=values.Length)
{
return false;
}
Dictionary<Char,String> dic=new Dictionary<Char,String>();
for(int i=;i<pattern.Length;i++)
{
if(!dic.ContainsKey(pattern[i]))
{
if (!dic.ContainsValue(values[i]))
{
dic.Add(pattern[i], values[i]);
}
else return false;
}
else
{
if(!dic[pattern[i]].Equals(values[i]))
{
return false;
}
}
}
return true;
}
}