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346 lines
11 KiB
Markdown
346 lines
11 KiB
Markdown
<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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> 在哈希法中有一些场景就是为数组量身定做的。
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# 383. 赎金信
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[力扣题目链接](https://leetcode-cn.com/problems/ransom-note/)
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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。
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(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。)
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**注意：**
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你可以假设两个字符串均只含有小写字母。
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canConstruct("a", "b") -> false
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canConstruct("aa", "ab") -> false
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canConstruct("aa", "aab") -> true
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## 思路
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这道题目和[242.有效的字母异位词](https://programmercarl.com/0242.有效的字母异位词.html)很像，[242.有效的字母异位词](https://programmercarl.com/0242.有效的字母异位词.html)相当于求 字符串a 和 字符串b 是否可以相互组成 ，而这道题目是求 字符串a能否组成字符串b，而不用管字符串b 能不能组成字符串a。
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本题判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成，但是这里需要注意两点。
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* 第一点“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” 这里*说明杂志里面的字母不可重复使用。*
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* 第二点 “你可以假设两个字符串均只含有小写字母。” *说明只有小写字母*，这一点很重要
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## 暴力解法
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那么第一个思路其实就是暴力枚举了，两层for循环，不断去寻找，代码如下：
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```CPP
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// 时间复杂度: O(n^2)
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// 空间复杂度：O(1)
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class Solution {
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public:
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bool canConstruct(string ransomNote, string magazine) {
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for (int i = 0; i < magazine.length(); i++) {
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for (int j = 0; j < ransomNote.length(); j++) {
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// 在ransomNote中找到和magazine相同的字符
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if (magazine[i] == ransomNote[j]) {
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ransomNote.erase(ransomNote.begin() + j); // ransomNote删除这个字符
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break;
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}
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}
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}
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// 如果ransomNote为空，则说明magazine的字符可以组成ransomNote
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if (ransomNote.length() == 0) {
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return true;
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}
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return false;
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}
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};
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```
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这里时间复杂度是比较高的，而且里面还有一个字符串删除也就是erase的操作，也是费时的，当然这段代码也可以过这道题。
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## 哈希解法
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因为题目所只有小写字母，那可以采用空间换取时间的哈希策略， 用一个长度为26的数组还记录magazine里字母出现的次数。
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然后再用ransomNote去验证这个数组是否包含了ransomNote所需要的所有字母。
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依然是数组在哈希法中的应用。
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一些同学可能想，用数组干啥，都用map完事了，**其实在本题的情况下，使用map的空间消耗要比数组大一些的，因为map要维护红黑树或者哈希表，而且还要做哈希函数，是费时的！数据量大的话就能体现出来差别了。 所以数组更加简单直接有效！**
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代码如下：
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```CPP
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// 时间复杂度: O(n)
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// 空间复杂度：O(1)
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class Solution {
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public:
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bool canConstruct(string ransomNote, string magazine) {
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int record[26] = {0};
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for (int i = 0; i < magazine.length(); i++) {
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// 通过recode数据记录 magazine里各个字符出现次数
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record[magazine[i]-'a'] ++;
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}
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for (int j = 0; j < ransomNote.length(); j++) {
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// 遍历ransomNote，在record里对应的字符个数做--操作
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record[ransomNote[j]-'a']--;
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// 如果小于零说明ransomNote里出现的字符，magazine没有
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if(record[ransomNote[j]-'a'] < 0) {
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return false;
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}
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}
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return true;
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}
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};
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```
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## 其他语言版本
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Java：
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```Java
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class Solution {
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public boolean canConstruct(String ransomNote, String magazine) {
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//记录杂志字符串出现的次数
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int[] arr = new int[26];
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int temp;
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for (int i = 0; i < magazine.length(); i++) {
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temp = magazine.charAt(i) - 'a';
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arr[temp]++;
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}
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for (int i = 0; i < ransomNote.length(); i++) {
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temp = ransomNote.charAt(i) - 'a';
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//对于金信中的每一个字符都在数组中查找
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//找到相应位减一，否则找不到返回false
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if (arr[temp] > 0) {
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arr[temp]--;
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} else {
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return false;
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}
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}
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return true;
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}
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}
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```
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Python写法一（使用数组作为哈希表）：
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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arr = [0] * 26
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for x in magazine:
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arr[ord(x) - ord('a')] += 1
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for x in ransomNote:
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if arr[ord(x) - ord('a')] == 0:
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return False
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else:
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arr[ord(x) - ord('a')] -= 1
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return True
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```
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Python写法二（使用defaultdict）：
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```python
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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from collections import defaultdict
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hashmap = defaultdict(int)
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for x in magazine:
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hashmap[x] += 1
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for x in ransomNote:
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value = hashmap.get(x)
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if value is None or value == 0:
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return False
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else:
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hashmap[x] -= 1
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return True
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```
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Python写法三：
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```python
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class Solution(object):
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def canConstruct(self, ransomNote, magazine):
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"""
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:type ransomNote: str
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:type magazine: str
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:rtype: bool
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"""
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# use a dict to store the number of letter occurance in ransomNote
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hashmap = dict()
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for s in ransomNote:
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if s in hashmap:
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hashmap[s] += 1
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else:
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hashmap[s] = 1
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# check if the letter we need can be found in magazine
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for l in magazine:
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if l in hashmap:
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hashmap[l] -= 1
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for key in hashmap:
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if hashmap[key] > 0:
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return False
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return True
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```
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Python写法四：
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```python3
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class Solution:
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def canConstruct(self, ransomNote: str, magazine: str) -> bool:
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c1 = collections.Counter(ransomNote)
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c2 = collections.Counter(magazine)
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x = c1 - c2
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#x只保留值大于0的符号，当c1里面的符号个数小于c2时，不会被保留
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#所以x只保留下了，magazine不能表达的
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if(len(x)==0):
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return True
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else:
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return False
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```
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Go：
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```go
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func canConstruct(ransomNote string, magazine string) bool {
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record := make([]int, 26)
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for _, v := range magazine {
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record[v-'a']++
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}
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for _, v := range ransomNote {
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record[v-'a']--
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if record[v-'a'] < 0 {
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return false
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}
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}
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return true
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}
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```
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javaScript:
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```js
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/**
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* @param {string} ransomNote
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* @param {string} magazine
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* @return {boolean}
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*/
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var canConstruct = function(ransomNote, magazine) {
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const strArr = new Array(26).fill(0),
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base = "a".charCodeAt();
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for(const s of magazine) {
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strArr[s.charCodeAt() - base]++;
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}
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for(const s of ransomNote) {
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const index = s.charCodeAt() - base;
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if(!strArr[index]) return false;
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strArr[index]--;
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}
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return true;
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};
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```
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PHP:
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```php
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class Solution {
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/**
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* @param String $ransomNote
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* @param String $magazine
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* @return Boolean
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*/
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function canConstruct($ransomNote, $magazine) {
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if (count($ransomNote) > count($magazine)) {
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return false;
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}
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$map = [];
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for ($i = 0; $i < strlen($magazine); $i++) {
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$map[$magazine[$i]] = ($map[$magazine[$i]] ?? 0) + 1;
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}
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for ($i = 0; $i < strlen($ransomNote); $i++) {
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if (!isset($map[$ransomNote[$i]]) || --$map[$ransomNote[$i]] < 0) {
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return false;
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}
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}
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return true;
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}
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```
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Swift：
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```swift
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func canConstruct(_ ransomNote: String, _ magazine: String) -> Bool {
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var record = Array(repeating: 0, count: 26);
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let aUnicodeScalarValue = "a".unicodeScalars.first!.value
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for unicodeScalar in magazine.unicodeScalars {
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// 通过record 记录 magazine 里各个字符出现的次数
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let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
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record[idx] += 1
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}
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for unicodeScalar in ransomNote.unicodeScalars {
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// 遍历 ransomNote,在record里对应的字符个数做 -- 操作
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let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
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record[idx] -= 1
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// 如果小于零说明在magazine没有
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if record[idx] < 0 {
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return false
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}
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}
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return true
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}
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```
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Rust:
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```rust
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impl Solution {
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pub fn can_construct(ransom_note: String, magazine: String) -> bool {
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let baseChar = 'a';
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let mut record = vec![0; 26];
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for byte in magazine.bytes() {
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record[byte as usize - baseChar as usize] += 1;
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}
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for byte in ransom_note.bytes() {
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record[byte as usize - baseChar as usize] -= 1;
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if record[byte as usize - baseChar as usize] < 0 {
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return false;
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}
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}
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return true;
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}
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}
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```
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-----------------------
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* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频：[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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