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
非常简单的一道题，就是用哈希Map统计字符的个数，参见代码如下：
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
unordered_map<char, int> m;
for (char c : magazine) ++m[c];
for (char c : ransomNote) {
if (--m[c] < 0) return false;
}
return true;
}
};
LeetCode All in One 题目讲解汇总(持续更新中...)
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.
非常简单的一道题，就是用哈希Map统计字符的个数，参见代码如下：
LeetCode All in One 题目讲解汇总(持续更新中...)