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383.ransom-note 赎金信
利用一个长度为26的数组记录magazine中每个字母出现的次数(递增)，再与ransom note中每个字母出现的次数进行对比(递减)即可。 #include <string> using std::string; class Solution { public: bool canConstruct(string ransomNote, string magazine) { int a[26] =LeetCode 383. Ransom Note
LeetCode 383. Ransom Note (赎金信) 题目 链接 https://leetcode.cn/problems/ransom-note/ 问题描述 给你两个字符串：ransomNote 和 magazine ，判断 ransomNote 能不能由 magazine 里面的字符构成。 如果可以，返回 true ；否则返回 false 。 magazine 中的每个字符只能在 ransomNote赎金信（Ransom Note）字符串比较的惯用套路
class Solution { public boolean canConstruct(String ransomNote, String magazine) { Boolean ans=true; int[] a=new int[26]; int[] b=new int[26]; for(int i=0;i<ransomNote.length();i++) { a[ransomNote.charAt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 ；
class Solution { public boolean canConstruct(String ransomNote, String magazine) { int[] letter1 = new int[123]; int[] letter2 = new int[123]; for (int i = 0; i < ransomNote.length(); i++){ letter1[ransom383. Ransom Note
Problem: 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[LeetCode]383. Ransom Note ★
每天一道编程题题目描述样例python解法C语言解法 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 ;