LeetCode 383. Ransom Note

题目：
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

思路：
先分别计算ransomNote和magazine的长度，如果ransomNote的长度大，直接返回false。如果ransomNote的长度大，直接返回false；如果在magazine中找到第一个ransomNode[i]，如果这个位置不是字符串中的最后（不过npos一般是-1），就将magazine中这个索引对应的字符删掉,继续循环，否则返回false。如果循环结束没有返回false，就可以构建，返回true。

代码：

class Solution {
public:bool canConstruct(string ransomNote, string magazine) {int lenran=ransomNote.length();//分别计算ransomNote和magazine的长度int lenmag=magazine.length();if(lenran>lenmag){//如果ransomNote的长度大，直接返回falsereturn false;}for(int i=0;i<lenran;++i){//遍历ransomNoteint index=magazine.find_first_of(ransomNote[i]);//如果在magazine中找到第一个ransomNode[i]if(index!=string::npos){//如果这个位置不是字符串中的最后（不过npos一般是-1）magazine.erase(index,1);//就将magazine中这个索引对应的字符删掉,继续循环continue;}else{//否则返回falsereturn false;}}return true;//如果循环结束没有返回false，就可以构建，返回true}
};

**输出结果：*42ms

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