题目要求:
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
原题链接:请戳这里
解决代码如下:
public class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
ArrayList<Character> ranList = getListFromArray(ransomNote);
ArrayList<Character> magList = getListFromArray(magazine);
// 循环从ransomNote中取出字符
for(int i=0;i<ranList.size();i++){
char c = ranList.get(i);
if(magList.contains(c)){
magList.remove(magList.indexOf(c));// *比对之后删掉该字符
continue;
}else{
return false;
}
}
return true;
}
// 将string转成list
public ArrayList<Character> getListFromArray(String string){
ArrayList<Character> list = new ArrayList<Character>();
char[] strArr = string.toCharArray();
for (char c : strArr) {
list.add(c);
}
return list;
}
}
我的思路:
1. 从ransomNote中逐个取出字符
2. 再到magazine中去确认是否含有该字符
3. 如果magazine中包含该字符,则删掉magazine中的这个字符
4. 如果magazine中不含该字符,则返回false
为什么要删掉magazine中的对应字符?因为假如说ransomNote为“aaagbfrd”,而magazine为“aabgefdrtg“时,针对字符‘a’就会出现错误判断,当判断ransomNote中第三个‘a’时,实际应该是false,但是magazine中还是有‘a’与之对应,故会返回错误值true。