思路,用unordered_map存儲magazine中字符以及字符出現的次數
遍歷ransomNote中每個字符,如果能在map中找到,則對應value減一,如果字符對應的value小于零,意味著magazine中找不到與ransomNote里這個字符對應的字符,返回false
/** @lc app=leetcode.cn id=383 lang=cpp** [383] 贖金信*/// @lc code=start
class Solution {
public:bool canConstruct(string ransomNote, string magazine) {// unordered_map<int,int> ransomNodeMap;unordered_map<char,int> magazineMap;for(char num:magazine){magazineMap[num]++;}for(char num:ransomNote){if(magazineMap[num] >0){magazineMap[num]--;}else{return false;}}return true;}
};
// @lc code=end