383. 赎金信

English Version

题目描述

给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。

(题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思。杂志字符串中的每个字符只能在赎金信字符串中使用一次。)

 

示例 1：

输入：ransomNote = "a", magazine = "b"
输出：false

示例 2：

输入：ransomNote = "aa", magazine = "ab"
输出：false

示例 3：

输入：ransomNote = "aa", magazine = "aab"
输出：true

 

提示：

• 你可以假设两个字符串均只含有小写字母。

解法

用一个数组或字典 chars 存放 magazine 中每个字母出现的次数。遍历 ransomNote 中每个字母，判断 chars 是否包含即可。

Python3

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        counter = Counter(magazine)
        for c in ransomNote:
            if counter[c] <= 0:
                return False
            counter[c] -= 1
        return True

Java

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] counter = new int[26];
        for (char c : magazine.toCharArray()) {
            ++counter[c - 'a'];
        }
        for (char c : ransomNote.toCharArray()) {
            if (counter[c - 'a'] <= 0) {
                return false;
            }
            --counter[c - 'a'];
        }
        return true;
    }
}

TypeScript

function canConstruct(ransomNote: string, magazine: string): boolean {
    let counter = new Array(26).fill(0);
    let base = "a".charCodeAt(0);
    for (let s of magazine) {
        ++counter[s.charCodeAt(0) - base];
    }
    for (let s of ransomNote) {
        let idx = s.charCodeAt(0) - base;
        if (counter[idx] == 0) return false;
        --counter[idx];
    }
    return true;
}

C++

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> counter(26);
        for (char c : magazine) ++counter[c - 'a'];
        for (char c : ransomNote)
        {
            if (counter[c - 'a'] <= 0) return false;
            --counter[c - 'a'];
        }
        return true;
    }
};

Go

func canConstruct(ransomNote string, magazine string) bool {
	rc := make([]int, 26)
	for _, b := range ransomNote {
		rc[b-'a']++
	}

	mc := make([]int, 26)
	for _, b := range magazine {
		mc[b-'a']++
	}

	for i := 0; i < 26; i++ {
		if rc[i] > mc[i] {
			return false
		}
	}
	return true
}

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