LeetCode--No.383--Ransom Note

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：

1. 对于字符串A中的每个字母，都去B中搜，搜到了就从B中删除。
但是空间复杂度比较高。O(m*n)，m为A字符串的长度，n为B字符串长度。
2. 空间换时间。
将字符串A中出现的每个字母，都记录一下次数。看看是不是比B中的次数少。
时间复杂度：O(max(m,n))

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] a = new int[26];
        int[] b = new int[26];
        for(int i = 0; i < ransomNote.length(); i++){
            char tmp = ransomNote.charAt(i);
            int num = tmp-'a';
            a[num]++;
        }
        for(int i = 0; i < magazine.length(); i++){
            char tmp = magazine.charAt(i);
            int num = tmp-'a';
            b[num]++;
        }
        for(int i = 0; i < 26; i++){
            if (a[i] > b[i])
                return false;
        }
        return true;
    }
}

原文链接：https://blog.csdn.net/sophia_tone2w/article/details/52202741