Word Pattern pattern = "abba", str = "dog cat cat dog" should return true

最新推荐文章于 2024-05-06 20:29:23 发布

chenyx90

最新推荐文章于 2024-05-06 20:29:23 发布

阅读量992

点赞数 2

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/chenyx90/article/details/50378562

leetcode 专栏收录该内容

57 篇文章

订阅专栏

本文探讨了如何通过模式匹配来验证字符串是否遵循特定模式，包括实现细节与常见案例分析。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 inpattern 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.

public class Solution {
    public boolean wordPattern(String pattern, String str) {
    String[] words = str.split(" ");
    if (words.length != pattern.length())
        return false;
    Map index = new HashMap();//map是个接口，实现需要hashmap以及treemap等
    for (Integer i=0; i<words.length; ++i)//这儿写成Integer i=0就对了，写成int i=0就出错了
        if (index.put(pattern.charAt(i), i)!=index.put(words[i], i))//put的返回值是return the previous value associated with key
            return false;
    return true;

    }
}

如果想要用Int则可以写成这样
for (int i=0; i<words.length; ++i)
if (!Objects.equals(index.put(pattern.charAt(i), i),index.put(words[i], i)))
return false;

==是判断两个变量或实例是不是指向同一个内存空间
equals是判断两个变量或实例所指向的内存空间的值是不是相同
除了String和封装器，equals()和“==”没什么区别
但String和封装器重写了equals()，所以在这里面，equals()指比较字符串或封装对象对应的原始值是否相等，"=="是比较两个对象是否为同一个对象
==是判断两个对象是否是同一个对象
equals是进行值的判断
String a = new String("aaa");
String b = new String("a");
b += "aa";
则 a==b //错误
a.equals(b)//正确

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] strs = str.split(" ");
        HashMap<String, Integer> mapstr = new HashMap<String, Integer>();
        HashMap<Character, Integer> mapc = new HashMap<Character, Integer>();
        if(pattern.length() != strs.length) return false;
        for(Integer i= 0;i < pattern.length();++i){
            char tmpc = pattern.charAt(i);
            String tmpstr = strs[i];//下一行的异或操作，不仅可以针对数字还可以针对逻辑
            if((mapc.containsKey(tmpc) ^ mapstr.containsKey(tmpstr))
                ||(mapc.containsKey(tmpc) && mapstr.containsKey(tmpstr) && mapc.get(tmpc) != mapstr.get(tmpstr)))
                return false;
            mapc.put(tmpc, i);//这个写法很巧妙的，算是变相的实现了上边的Put对比的那种写法
            mapstr.put(tmpstr, i);//注意一定要把这两句Put放在if语句的后边，如果放在前边那完全就坏了
        }
        return true;
    }
}