3. Then, traverse`ransomNote` and perform reverse operations on the data in`Map`. If the count of a character is less than 0, return`false`.

1. First count the characters in`magazine`, and store the results in`Map`.

charToCount=newMap()

for (characterin magazine) {

charToCount[character]+=1

}

2. Then, traverse`ransomNote` and perform reverse operations on the data in`Map`. If the count of a character is less than 0, return`false`.

charToCount=newMap()

for (characterin magazine) {

charToCount[character]+=1

}

for (characterin ransomNote) {

charToCount[character]-=1

if (charToCount[character]<0) {

}

}

* Time:`O(n)`.

* 线段树https://leetcode.cn/problems/fruits-into-baskets-iii

LeetCode link:[383. Ransom Note](https://leetcode.com/problems/ransom-note),

[383. 赎金信](https://leetcode.cn/problems/ransom-note)

原文链接：[coding5.com - 力扣题解最佳实践](https://coding5.com/zh/leetcode/383-ransom-note)

Given two strings`ransomNote` and`magazine`, return`true` if`ransomNote` can be constructed by using the letters from`magazine` and`false` otherwise.

力扣链接：[383. 赎金信](https://leetcode.cn/problems/ransom-note), 难度：**简单**。

**Input**:`ransomNote = "a", magazine = "b"`

如果可以，返回`true` ；否则返回`false` 。

**Input**:`ransomNote = "aa", magazine = "ab"`

**Output**:`false`

**输入**:`ransomNote = "a", magazine = "b"`

**Output**:`true`

**输入**:`ransomNote = "aa", magazine = "ab"`

-`ransomNote` and`magazine` consist of lowercase English letters.

**输出**:`false`

[中文题解](#中文题解)

**输入**:`ransomNote = "aa", magazine = "aab"`

1. This question is equivalent to asking whether`magazine` can contain all the characters in`ransomNote`.

2. First count the characters in`magazine`, and store the results in`Map`.

3. Then, traverse`ransomNote` and perform reverse operations on the data in`Map`. If the count of a character is less than 0, return`false`.

**输出**:`true`

1. First count the characters in`magazine`, and store the results in`Map`.

charToCount=newMap()

-`ransomNote` 和`magazine` 由小写英文字母组成

for (characterin magazine) {

charToCount[character]+=1

}

2. Then, traverse`ransomNote` and perform reverse operations on the data in`Map`. If the count of a character is less than 0, return`false`.

charToCount=newMap()

1. 本题等同于求`magazine`是否能包含`ransomNote`中的所有字符。

2. 先对`magazine`进行统计，得出每个字符对应的字数，结果存储在`Map`中。每一次都是一个加一的操作。

3. 下一步做什么？

<details><summary>点击查看答案</summary><p>遍历`ransomNote`，对当前字符对应的数量进行减一操作（反向操作）。如果某个字符的数量小于0，则返回`false`。</p></details>

for (characterin magazine) {

charToCount[character]+=1

}

for (characterin ransomNote) {

charToCount[character]-=1

1. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。

if (charToCount[character]<0) {

}

}

2. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。

* Time:`O(n)`.

* Space:`O(n)`.

- 时间复杂度:`O(N)`.

- 空间复杂度:`O(N)`.

classSolution {

publicbooleancanConstruct(StringransomNote,Stringmagazine) {

varcanConstruct=function (ransomNote,magazine) {

constcharToCount=newMap()

{

}

[383. 赎金信](https://leetcode.cn/problems/ransom-note) ，难度：**简单**。

给你两个字符串：`ransomNote` 和`magazine` ，判断`ransomNote` 能不能由`magazine` 里面的字符构成。

如果可以，返回`true` ；否则返回`false` 。

`magazine` 中的每个字符只能在`ransomNote` 中使用一次。

**输入**:`ransomNote = "aa", magazine = "ab"`

**输出**:`false`

**输入**:`ransomNote = "aa", magazine = "aab"`

**输出**:`true`

1. 本题等同于求`magazine`是否能包含`ransomNote`中的所有字符。

2. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。

3. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。

1. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。

charToCount=newMap()

for (characterin magazine) {

charToCount[character]+=1

}

2. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。

charToCount=newMap()

for (characterin magazine) {

charToCount[character]+=1

}

for (characterin ransomNote) {

charToCount[character]-=1

if (charToCount[character]<0) {

}

}