leetcode [#383]

目录

• 1. 题目
• 2. 解决方案
• 3. 注意事项

题目

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.

Example:
canConstruct(“a”, “b”) -> false
canConstruct(“aa”, “ab”) -> false
canConstruct(“aa”, “aab”) -> true

Note:
You may assume that both strings contain only lowercase letters.

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解决方案

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int len1 = ransomNote.length();
        int len2 = magazine.length();
        if(len1 > len2) return false;
        boolean result = true;
        List<String> list = new ArrayList<>();
        for(int k = 0; k < len2; k++){
            list.add(String.valueOf(magazine.charAt(k)));
        }
        for(int i = 0; i < len1; i++){
            if(!list.remove(String.valueOf(ransomNote.charAt(i)))){
                result = false;
                break;
            }
        }
        return result;
    }
}

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注意事项

1. 先判断特殊情况，如果magazine的长度比ransomNote小，那么一定不能组成结果。
2. 将magazine的每个字符存入list备用。
3. 遍历ransomNote，将每个字符从list中弹出，如果成功，证明list中有这个字符，也就是能作为构成ransomNote的一个字符，如果弹出失败，说明没有这个字符，返回false即可。