Ransom Note

标签：

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

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] arr = new int[26];
        for(int i=0;i<magazine.length();i++)
        {
            char c = magazine.charAt(i);
            arr[c-‘a‘]++;
        }
        
        for(int i=0;i<ransomNote.length();i++)
        {
            char c = ransomNote.charAt(i);
            arr[c-‘a‘]--;
            if(arr[c-‘a‘]<0) return false;
        }
        
        return true;
    }
}

reference: 

https://discuss.leetcode.com/topic/53864/java-o-n-solution-easy-to-understand/3

Ransom Note

标签：

原文地址：http://www.cnblogs.com/hygeia/p/5769389.html