290 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 returntrue
. - pattern =
"abba"
, str ="dog cat cat fish"
should returnfalse
. - pattern =
"aaaa"
, str ="dog cat cat dog"
should returnfalse
. - pattern =
"abba"
, str ="dog dog dog dog"
should returnfalse
.
Notes:
You may assume pattern
contains only lowercase letters, and str
contains lowercase letters separated by a single space.
分析& 代码
哈哈这题算是我splunk面试的题目了
可以建两个map,一个正向记录map
什么长度不同的废话不说了,核心逻辑是:
如果map里面没有patternLetter
那么如果revMap有word[i],返回false
不然把这一组正反加入两个map
否则
如果map对应的word和当前不和 || revMap没有当前word || revMap对应pattern不是当前pattern
返回false
返回true
代码:
public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (words.length != pattern.length())
return false;
Map index = new HashMap();
for (Integer i=0; i<words.length; ++i)
if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
return false;
return true;
}
但是呢
大家都讨厌两个map,用一个map也可以做,只要知道有一个containsValue()函数就可以了,好像是在没啥要解释的
如果已经有的话,就比较一下存的和现在一不一样
如果没有的话,就看下现在的word有没有在value里
public boolean wordPattern(String pattern, String str) {
if(pattern == null || str == null) {
return false;
}
String[] words = str.split(" ");
if(words.length != pattern.length()) {
return false;
}
Map<Character, String> map = new HashMap<>();
for(int i = 0; i < pattern.length(); i++) {
char c = pattern.charAt(i);
if(map.containsKey(c)) {
if(!map.get(c).equals(words[i])) {
return false;
}
} else {
if(map.containsValue(words[i])) {
return false;
}
map.put(c, words[i]);
}
}
return true;
}
Steven大神依旧有神奇的做法
public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (words.length != pattern.length())
return false;
Map index = new HashMap();
for (Integer i=0; i<words.length; ++i)
if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
return false;
return true;
}
其实我觉得这种做法不是那么好啦,因为map都没有说明类型,虽然都是Object类型,但是感觉不是很好呢……
这里要说的意思是,发现了map函数的神奇特质
就是map.put(),如果第二次put进同样的key不同value,返回的值是被挤掉的那个value!不然就是null
所以大神的代码的意思是如果下一次发现了一对Pattern+word,他们上次出现的位置应该是一样的。就是现在放入的这一对,如果在map中已经存在,那么上次放入的也应该是同一对