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 题目链接
和LintCode55题一个意思嘛，就是一旦第一个字符串的某个字符个数少于第二个字符串的相同字符个数，则false。
最easy的解法就是拿一个26大小的数组对应26个字母，先对第二个字符串取出每个字符，在数组对应的字符位置加加计数，再对第一个字符串减减计数，若其中有一个位置小于0，则说明第一个字符串有字符未曾在第二个字符串出现。连着LintCode刷完55又来刷这道题，改进了一下代码，没想到刚好就是discuss里的best solution
public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int[] count = new int[26]; for(int i = 0; i < magazine.length(); ++i) ++count[magazine.charAt(i) - 'a']; for(int i = 0; i < ransomNote.length(); ++i) if(--count[ransomNote.charAt(i) - 'a'] < 0) return false; return true; } }