Java [Leetcode 383]Ransom Note

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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 ?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

解题思路：

开26个数组存入magzine字母的数目，每个位置存对应字母的数目；

对ransom字符串来说，每读一个字母，则将对应位置的字母数目减一，如果某个字母数目小于0了，则表明字母不够用，从而返回false；否则返回true。

代码如下：

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
    	int[] ran_array = new int[26];
    	for(int i = 0; i < magazine.length(); i++){
    		ran_array[magazine.charAt(i) - ‘a‘]++;
    	}

    	for(int i = 0; i < ransomNote.length(); i++){
    		if(--ran_array[ransomNote.charAt(i) - ‘a‘] < 0)
    			return false;
    	}

    	return true;
        
    }
}

Java [Leetcode 383]Ransom Note

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原文地址：http://www.cnblogs.com/zihaowang/p/5778378.html