标签：you leetcode bsp etc OLE false href struct http
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
看到这道题目的时候，第一时间想到的是map来进行记录每个单词的数量，然后进行判断
但是看讨论区后，发现可以直接用数组来进行记录，很方便的思路啊！！！
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
if(ransomNote == null || ransomNote.length() == 0) return true;
if(magazine == null || magazine.length() == 0) return false;
char[] note = ransomNote.toCharArray();
char[] letter = magazine.toCharArray();
Map<Character, Integer> map = new HashMap<>();
for(int i = 0; i < letter.length; i++){
map.put(letter[i], map.getOrDefault(letter[i], 0) + 1);
}
for(int i = 0; i < note.length; i++){
Integer num = map.get(note[i]);
if(num != null && num >= 1){
map.put(note[i], num - 1);
} else{
return false;
}
}
return true;
}
}
数组的方式：
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
if(ransomNote == null || ransomNote.length() == 0) return true;
if(magazine == null || magazine.length() == 0) return false;
int[] letter = new int[26];
for(int i = 0; i < magazine.length(); i++){
letter[magazine.charAt(i) - ‘a‘]++;
}
for(int i = 0; i < ransomNote.length(); i++){
if(--letter[ransomNote.charAt(i) - ‘a‘] < 0) return false;
}
return true;
}
}
标签：you leetcode bsp etc OLE false href struct http
原文地址：https://www.cnblogs.com/SkyeAngel/p/9079340.html
