- Word Pattern
Easy
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.
Example 1:
Input: pattern = “abba”, str = “dog cat cat dog”
Output: true
Example 2:
Input:pattern = “abba”, str = “dog cat cat fish”
Output: false
Example 3:
Input: pattern = “aaaa”, str = “dog cat cat dog”
Output: false
Example 4:
Input: pattern = “abba”, str = “dog dog dog dog”
Output: false
Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
给定一种 pattern(模式) 和一个字符串 str ，判断 str 是否遵循相同的模式。
这里的遵循指完全匹配，例如， pattern 里的每个字母和字符串 str 中的每个非空单词之间存在着双向连接的对应模式。
示例1:
输入: pattern = “abba”, str = “dog cat cat dog”
输出: true
示例 2:
输入:pattern = “abba”, str = “dog cat cat fish”
输出: false
示例 3:
输入: pattern = “aaaa”, str = “dog cat cat dog”
输出: false
示例 4:
输入: pattern = “abba”, str = “dog dog dog dog”
输出: false
说明:
你可以假设 pattern 只包含小写字母， str 包含了由单个空格分隔的小写字母。
解题思路：此题和205同构字符串是同一个套路。
理由hash表，把字符和字符串相互映射，然后判断。
代码：
class Solution {
public:
bool wordPattern(string pa, string str) {
if(pa.size() ==0 || str.length() == 0 ) return false;
unordered_map < char ,string > ch;
unordered_map < string ,char> s;
vector <string> re(pa.size());
int j=0;
for( int i = 0; i < pa.size(); i++)
{
if(j>=str.length())return false; // pattern字符数比str中的单词数多。
while( j < str.length() ) //把某个单词读取出来
{
if (str[j] != ' ')
{
re[i]+=str[j];
j++;
}
else
{
j++;
break;
}
}
if(ch.count(pa[i]) && s.count(re[i]) )
{
if(ch[pa[i]] != re[i])
return false;
}
else if (ch.count(pa[i]) || s.count(re[i]) )
return false;
else {
ch[pa[i]] = re[i];
s[re[i]] = pa[i];
}
}
if (j < str.length()) return false; //str中的单词数比pattern中的字符多。
return true;
}
};