Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

Note:
You may assume both and have the same length.

Accepted solution:

create an array: map all the character and count the value as the index of each characters in string. start from 1

e.g.: egg & dda & add

a['e'] = 1 a['g'] = 2 -> 3 ; a['d'] = 1 -> 2  a['a'] 3 ; a['a'] = 1   a['d'] 2 -> 3

class Solution {
public boolean isIsomorphic(String s, String t) {
//letter , index of string
//e.g. egg : e: 1 g:2 g:3
if(s.length()!=t.length()) return false;
int[] a = new int[256];
int[] b = new int[256];
for(int i = 0; i<s.length(); i++){
if(a[s.charAt(i)] != b[t.charAt(i)]) return false;
//update the character
a[s.charAt(i)] = i+1; // why plus 1 for the case aa & ab
b[t.charAt(i)] = i+1;
}
return true;
}
}

Good idea make the code clean and easy. Think before coding.

*205. Isomorphic Strings (string & map idea)的更多相关文章

  1. 205. Isomorphic Strings - LeetCode

    Question 205. Isomorphic Strings Solution 题目大意:判断两个字符串是否具有相同的结构 思路:构造一个map,存储每个字符的差,遍历字符串,判断两个两个字符串中 ...

  2. [leetcode]205. Isomorphic Strings 同构字符串

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  3. 【刷题-LeetCode】205. Isomorphic Strings

    Isomorphic Strings Given two strings *s* and *t*, determine if they are isomorphic. Two strings are ...

  4. LeetCode 205 Isomorphic Strings(同构的字符串)(string、vector、map)(*)

    翻译 给定两个字符串s和t,决定它们是否是同构的. 假设s中的元素被替换能够得到t,那么称这两个字符串是同构的. 在用一个字符串的元素替换还有一个字符串的元素的过程中.所有字符的顺序必须保留. 没有两 ...

  5. (String) 205.Isomorphic Strings

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  6. 205. Isomorphic Strings (Map)

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  7. LeetCode 205 Isomorphic Strings

    Problem: Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if ...

  8. Java for LeetCode 205 Isomorphic Strings

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

  9. 205 Isomorphic Strings

    Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the chara ...

随机推荐

  1. eval()解析json以及js中js数组、对象与json之间的转换

    http://www.cnblogs.com/myjavawork/articles/1979279.html https://www.cnblogs.com/coder-economy/p/6203 ...

  2. Oracle 树形SQL语句,SYS_CONNECT_BY_PATH 函数

    转一个SYS_CONNECT_BY_PATH 函数的例子.推断原表应该是这样: Child                        Parent ------------------------ ...

  3. RequireJS -Javascript模块化(二、模块依赖)

    上一篇文章中简单介绍了RequireJs的写法和使用,这节试着写下依赖关系 需求描述:我们经常写自己的js,在元素选择器这方面,我们可能会用jquery的$("#id")id选择器 ...

  4. oracle merge into函数中插入clob字段

    当使用Merge into 函数向ORACLE数据库中插入或更新数据时,报错“ORA-01461: 仅可以为插入 LONG 列的 LONG 值赋值”,使用如下方法可以把String转换为clob. 需 ...

  5. Java 实践

    /** *宠物就是一个标准,包含多类宠物 *定义宠物标准接口Pet *定义Cat和Dog两个Pet接口的子类 *使用链表结构动态存储宠物信息 *定义商店类(工厂类),负责宠物的上架(链表添加).下架( ...

  6. equals和等号的区别

    如果是基本类型,等号比较的是数值.如果是引用类型,等号比较的是地址.而equals如果没有重写的话默认比较的是地址,可以重写equals来自定义比较两个对象的逻辑.

  7. Cookie和Session入门(一)

    目录一)背景介绍二)Cookie机制三)Session机制四)两者比较五)参考资料链接一)背景介绍Cookie与Session是常用的会话跟踪技术.1.Cookie通过在客户端记录信息确定用户身份,S ...

  8. 多线程编程_控制并发线程数的Semaphore

    简介 Semaphore(信号量)是用来控制同时访问特定资源的线程数量,它通过协调各个线程,以保证合理的使用公共资源.很多年以来,我都觉得从字面上很难理解Semaphore所表达的含义,只能把它比作是 ...

  9. RTT设备与驱动之SPI

    SPI全双工设备的操作分为主设备和从设备(可以多个,多线程下从设备访问主设备要先获得总线控制权) rt_device_t rt_device_find(const char* name);查找设备 s ...

  10. STM32 从M3到M4

    一 考虑STM32不同系列移植的外设资源情况: STM32微控制器应用的移植和兼容性指南AN3364 二 M4的DSP/FPU的使用方法https://blog.csdn.net/electrocra ...