LeetCode oj 383. Ransom Note(雜湊）
阿新 • • 發佈：2019-01-10
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 magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
給你兩個字串，問第一個字串能不能由第二個字串中的字元組成
水的不行，雜湊存一下就行了。
public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int hash_r[] = new int [256]; int hash_m[] = new int [256]; int len_r = ransomNote.length(); int len_m = magazine.length(); if(len_r > len_m) return false; for(int i=0;i<len_r;i++){ hash_r[ransomNote.charAt(i)-'a']++; }; for(int i=0;i<len_m;i++){ hash_m[magazine.charAt(i)-'a']++; } int flag = 0; for(int i=0;i<len_r;i++){ if(hash_m[ransomNote.charAt(i)-'a'] < hash_r[ransomNote.charAt(i)-'a']){ flag = 1; break; } } if(flag == 1) return false; else return true; } }