leetcode383. Ransom Note（赎金信）JAVA实现

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

实现思路：定义两个辅助数组temp1和temp2，用来记录ransomNote和magazine中的各个字母个数，然后检查如果ransomNote有需要使用某个字母m次，则检查magazine中是否存在对应的字母且个数大于等于m。

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int []temp1 = new int[26];
        int []temp2 = new int[26];
        for(int i=0;i<ransomNote.length();i++){
            temp1[ransomNote.charAt(i)-'a']++;
        }
        for(int i=0;i<magazine.length();i++){
            temp2[magazine.charAt(i)-'a']++;
        }
        for(int i=0;i<26;i++){
            if(temp1[i]!=0){
                if(temp1[i]>temp2[i])
                    return false;
            }
        }
        return true;
    }
}

实现代码二：使用一个辅助数组也可以

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        boolean ans=true;
        int[] marked=new int[256];
        char[] c=magazine.toCharArray();
        char[] x=ransomNote.toCharArray();
        for(int i=0; i<c.length; i++){
            marked[c[i]]++;
        }
        for(int i=0; i<x.length; i++){
            if(marked[x[i]]==0){
                ans=false;
                break;
            }
            else{
                marked[x[i]]--;
            }
        }
        return ans;
    }
}