LeetCode:Ransom Note_383】的更多相关文章

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…

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…
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…
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.…
-------------------------------------------- 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…

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…
#-*- 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))…
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…
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…
这是悦乐书的第212次更新,第225篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第80题(顺位题号是383).给定一个任意赎金票据字符串和另一个包含所有杂志字母的字符串,如果赎金票据可以从杂志中构建,则写一个函数将返回true;否则,它将返回false.杂志字符串中的每个字母只能在赎金票据中使用一次.例如: canConstruct("a","b") - > false canConstruct("aa",&…
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…
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…
题目描述: 
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…
题目描述: 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…
题目: 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…
终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…
Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…
自从上个月进入实验室的云安全项目组后,因为要接触到实际的代码,在实验室博士的建议下我们项目组的硕士开始刷LeetCode练习编程能力,保持每周抽空刷几道算法题.虽然刷的不多,到现在一共只刷了不到30题,但在刷题的过程中还是有很多感触的. 实验室的博士建议我们按照题目的难易顺序循序渐进由易到难来刷,我也就照做了.因为前段时间在系统地学Python语言,所以我主要用的是Python语言来做,有的题目也采用了C/C++.Java甚至C#多种语言来尝试.在刷题的过程中,总结了自己的一些问题: 1. 对于…
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…
刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…
We are given N different types of stickers. Each sticker has a lowercase English word on it. You would like to spell out the given target string by cutting individual letters from your collection of stickers and rearranging them. You can use each sti…
请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…
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…
刷题备忘录,for bug-free 招行面试题--求无序数组最长连续序列的长度,这里连续指的是值连续--间隔为1,并不是数值的位置连续 问题: 给出一个未排序的整数数组,找出最长的连续元素序列的长度. 如: 给出[100, 4, 200, 1, 3, 2], 最长的连续元素序列是[1, 2, 3, 4].返回它的长度:4. 你的算法必须有O(n)的时间复杂度 . 解法: 初始思路 要找连续的元素,第一反应一般是先把数组排序.但悲剧的是题目中明确要求了O(n)的时间复杂度,要做一次排序,是不能达…
突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…
344. Reverse String Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". Subscribe to see which companies asked this question public class Solution { public String r…
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.canConstruct("a", "b") -> falsecanConstruct("aa",…
All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…
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源代码地址:https://github.com/hopebo/hopelee 语言:C++ 301. Remove Invalid Parentheses Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. Note: The input string may contain letters other tha…