Algorithm-Word Law
Word law
给定一种规律 pattern 和一个字符串 str ，判断 str 是否遵循相同的规律。
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规律。
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。
The difficulty of this topic is easy, but it is very distinctive. It is a typical example of examining the idea of hash.
How to solve the problem? We can match two strings by establishing an intermediate mapping relationship
In this question, we can map the character in the pattern to the index of its first appearance according to its first appearance position, which is
abba->0110
At the same time, the str is mapped according to similar rules
dog cat cat fish->0110
At this time, we only need to compare whether the values of the mapping between them are the same,
but when we perform the second mapping, since we already have the value of the first mapping, we don’t need to open an array to store the second one. After mapping, we can directly compare the value of the corresponding bit.
//优化后的，击败98%
public boolean wordPattern(String pattern, String str) {
Map<Character,Integer> cmap=new HashMap<>();//记录元素出现的第一个位置
char[] cp=pattern.toCharArray();
String[] cs=str.split(" ");
if(cs.length!=cp.length){
return false;
}
int[] ccp=new int[pattern.length()];
for(int i=0;i<cp.length;i++){
if(!cmap.containsKey(cp[i])){
cmap.put(cp[i],i);
}
ccp[i]=cmap.get(cp[i]);
}
Map<String,Integer> smap=new HashMap<>();//记录元素出现的第一个位置
for(int i=0;i<cs.length;i++){
if(!smap.containsKey(cs[i])){
smap.put(cs[i],i);
}
if(ccp[i]!=smap.get(cs[i])){
return false;
}
}
return true;
}
//未优化的，击败8%
public boolean wordPattern(String pattern, String str) {
Map<Character,Integer> cmap=new HashMap<>();//记录元素出现的第一个位置
char[] cp=pattern.toCharArray();
StringBuilder sb1=new StringBuilder();
for(int i=0;i<cp.length;i++){
if(!cmap.containsKey(cp[i])){
cmap.put(cp[i],i);
}
sb1.append(cmap.get(cp[i]));
}
String[] cs=str.split(" ");
Map<String,Integer> smap=new HashMap<>();//记录元素出现的第一个位置
StringBuilder sb2=new StringBuilder();
for(int i=0;i<cs.length;i++){
if(!smap.containsKey(cs[i])){
smap.put(cs[i],i);
}
sb2.append(smap.get(cs[i]));
}
if(sb1.toString().equals(sb2.toString())){
return true;
}
return false;
}