LeetCode:Ransom Note_383

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LeetCode:Ransom Note_383

LeetCode:Ransom Note [问题再现] Given  an  arbitrary  ransom  note  string  and  another  string  containing  letters from  all  the  magazines,  write  a  function  that  will  return  true  if  the  ransom   note  can  be  constructed  from  the  magaz…

[LeetCode] Ransom Note 赎金条

 Given  an  arbitrary  ransom  note  string  and  another  string  containing  letters from  all  the  magazines,  write  a  function  that  will  return  true  if  the  ransom   note  can  be  constructed  from  the  magazines ;  otherwise,  it  wil…

LeetCode: Ransom Note

public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int[] ransomNum = new int[256]; int[] magNum = new int[256]; for (int i = 0; i < 256; i++) { ransomNum[i] = magNum[i] = 0; } for (int i = 0; i < ransomNote.len…

C#LeetCode刷题之#383-赎金信（Ransom Note）

问题该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3937 访问. 给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串,判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成.如果可以构成,返回 true :否则返回 false. (题目说明:为了不暴露赎金信字迹,要从杂志上搜索各个需要的字母,组成单词来表达意思.) 注意:你可以假设两个字符串均只含有小写字母.…

【LeetCode】383. Ransom Note 解题报告(Java & Python)

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录题目描述题目大意解题方法 Java解法 Python解法日期 [LeetCode] 题目地址:https://leetcode.com/problems/ransom-note/ Difficulty: Easy 题目描述 Given an arbitrary ransom note string and another string containing letters…

LeetCode之383. Ransom Note

-------------------------------------------- 思路就是进行频率统计. 统计一下第二个字符串字符出现次数++统计一下第一个字符串中字符出现次数--如果出现负数说明第二个中的字符不够用的. AC代码如下: public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int book[]=new int[26]; for(int i=0;i…

leetcode 383. Ransom Note

 Given  an  arbitrary  ransom  note  string  and  another  string  containing  letters from  all  the  magazines,  write  a  function  that  will  return  true  if  the  ransom   note  can  be  constructed  from  the  magazines ;  otherwise,  it  wil…

【leetcode❤python】Ransom Note

#-*- coding: UTF-8 -*- class Solution(object): def canConstruct(self, ransomNote, magazine): ransomNote=list(ransomNote) magezine=list(magazine) for ransom in ransomNote: if(magezine.__contains__(ransom))…

leetcode修炼之路——383. Ransom Note

题目是这样的 Given  an  arbitrary  ransom  note  string  and  another  string  containing  letters from  all  the  magazines,  write  a  function  that  will  return  true  if  the  ransom   note  can  be  constructed  from  the  magazines ;  otherwise,  i…

14. leetcode 383. Ransom Note

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false. Each letter in the…