[刷题防痴呆] 0383 - 赎金信 (Ransom Note)

题目地址

https://leetcode.com/problems/ransom-note/description/

题目描述

383. Ransom Note

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 1:

Input: ransomNote = "a", magazine = "b"
Output: false
Example 2:

Input: ransomNote = "aa", magazine = "ab"
Output: false
Example 3:

Input: ransomNote = "aa", magazine = "aab"
Output: true
 

Constraints:

You may assume that both strings contain only lowercase letters.

思路

• 用一个count数组.
• 第一遍遍历magazine的内容记录每个字符出现的个数.
• 第二遍遍历ansomNote的内容减去相应字符的个数, 若出现某个字符的个数小于0, 则返回false.

关键点

代码

• 语言支持：Java

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

// hashmap
class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c: magazine.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        for (char c: ransomNote.toCharArray()) {
            if (!map.containsKey(c)) {
                return false;
            }
            if (map.get(c) == 0) {
                return false;
            }
            map.put(c, map.get(c) - 1);
        }

        return true;
    }
}