leetcode 383. Ransom Note HashMap 统计字符 + HashMap

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.

Note:

You may assume that both strings contain only lowercase letters.

canConstruct(“a”, “b”) -> false

canConstruct(“aa”, “ab”) -> false

canConstruct(“aa”, “aab”) -> true

直接使用HashMap遍历即可。

代码如下：

import java.util.HashMap;
import java.util.Map;

class Solution
{
public boolean canConstruct(String ransomNote, String magazine)
{
if(ransomNote==null || magazine==null || ransomNote.length()>magazine.length())
return false;

Map<Character, Integer> map=new HashMap<>();
for(int i=0;i<magazine.length();i++)
map.put(magazine.charAt(i), map.getOrDefault(magazine.charAt(i), 0)+1);
for(int i=0;i<ransomNote.length();i++)
{
if(map.containsKey(ransomNote.charAt(i))==false)
return false;
else
{
int size=map.get(ransomNote.charAt(i));
if(size==1)
map.remove(ransomNote.charAt(i));
else
map.put(ransomNote.charAt(i), size-1);
}
}
return true;
}
}

下面是C++的做法，自己利用一个数组做统计即可

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

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