【Leetcode】Word Pattern

题目链接：https://leetcode.com/problems/word-pattern/

题目：

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.

思路：

1、pattern字符相等的位置对应的字符串中的单词应该是相等的，pattern字符不相等的位置对应的字符串中单词应该也是不相等的。时间复杂度为

O(n^2)，空间复杂度为O(1)。

2、用HashMap存储字符到单词的映射关系，时间复杂度为O(n)，空间复杂度为O(n)。

算法1：

[java] view
plain copy







public boolean wordPattern(String pattern, String str) {

String strs[] = str.split(" ");

char patterns[] = pattern.toCharArray();

if (strs.length != patterns.length || strs.length == 0) {

return false;

}

for (int i = 0; i < patterns.length; i++) {

for (int j = i + 1; j < patterns.length; j++) {

if (patterns[i] == patterns[j]) { // 字符相等的时候，单词应该相等

if (!strs[i].equals(strs[j])) {

return false;

}

} else { // 字符不相等的时候，单词也应该不相等

if (strs[i].equals(strs[j])) {

return false;

}

}

}

}

return true;

}

算法2：

[java] view
plain copy







public boolean wordPattern(String pattern, String str) {

String strs[] = str.split(" ");

char patterns[] = pattern.toCharArray();

HashMap<Character,String> map = new HashMap<Character,String>();

HashMap<String,Character> map2 = new HashMap<String,Character>();

if (strs.length != patterns.length || strs.length == 0) {

return false;

}

for (int i = 0; i < patterns.length; i++) {

if(!map.containsKey(patterns[i])){

map.put(patterns[i], strs[i]);

}else{

if(!map.get(patterns[i]).equals(strs[i])){

return false;

}

}

if(!map2.containsKey(strs[i])){

map2.put(strs[i], patterns[i]);

}else{

if(!map2.get(strs[i]).equals(patterns[i])){

return false;

}

}

}

return true;

}