[LeetCode] Word Pattern

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.

解题思路

用一个哈希表建立从pattern中字符到str中字符串的映射。

用一个集合表示str中字符串是否已作为值出现在哈希表中。

以下两种情况为假：

哈希表中已存在key，但是对应的value值不一样。

哈希表中不存在key，但是对应的值已作为其他key的value值。

实现代码

C++:

// Runtime: 0 ms
class Solution {
public:
bool wordPattern(string pattern, string str) {
vector<string> words;
istringstream iss(str);
string temp;
while (iss>>temp)
{
words.push_back(temp);
}
if (words.size() != pattern.size())
{
return false;
}
unordered_map<char, string> mymap;
set<string> used;
for (int i = 0; i < pattern.length(); i++)
{
if (mymap.find(pattern[i]) == mymap.end())
{
if (used.find(words[i]) == used.end())
{
mymap[pattern[i]] = words[i];
used.insert(words[i]);
}
else
{
return false;
}
}
else if (mymap[pattern[i]].compare(words[i]) != 0)
{
return false;
}
}

return true;
}
};

Java:

// Runtime: 3 ms
public class Solution {
public boolean wordPattern(String pattern, String str) {
String words[] = str.split(" ");
if (pattern.length() != words.length) {
return false;
}
Map<Character, String> mymap = new HashMap<Character, String>();
for (int i = 0; i < pattern.length(); i++) {
if (!mymap.containsKey(pattern.charAt(i))) {
if (!mymap.containsValue(words[i])) {
mymap.put(pattern.charAt(i), words[i]);
}
else {
return false;
}
}
else if (!mymap.get(pattern.charAt(i)).equals(words[i])) {
return false;
}
}

return true;
}
}