LeetCode之Ransom Note (Java+C)

题意：
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") -> falsecanConstruct("aa", "ab") -> falsecanConstruct("aa", "aab") -> true

题意确实很绕，表示没看明白，后来上网翻译了下才理解了，可以很简单的理解为：前面的字符串ransomNote能否完全由后面的字符串magazine的字符所组成（都只考虑为小写字母情况）

Java代码：

public class Solution {    public boolean canConstruct(String ransomNote, String magazine) {        char[] ransoms = ransomNote.toCharArray();        char[] magazines = magazine.toCharArray();        for(int i = 0; i < ransoms.length; i++) {            for(int j = 0; j < magazines.length; j++) {                if(ransoms[i] == magazines[j]) {                    ransoms[i] = 0;                    magazines[j] = 0;                    break;                }            }        }        for(char r : ransoms) {            if(r != 0) return false;        }        return true;    }}

C代码：

bool canConstruct(char* ransomNote, char* magazine) {    int ransomLen = sizeof(ransomNote);    int magazineLen = sizeof(magazine);    for(int i = 0; i < ransomLen; i++) {        for(int j = 0; j < magazineLen; j++) {            if(ransomNote[i] == magazine[j]) {                ransomNote[i] = 0;                magazine[j] = 0;                break;            }        }    }    for(int i = 0; i < ransomLen; i++) {        if(ransomNote[i] != 0)            return false;    }    return true;}

解题思路：遍历数组，一一比对，相同的话，相应两数组的元素均置零！最后重新遍历ransomNote数组，如果存在非零元素，说明不符合要求，返回false，反之返回true