LeetCode Ransom Note 383

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

自己第一次做 LeetCode 上面的题，也没什么算法方面的学习，摸着石头过河，步步提高。

大体意思

判断第二个字符串中的字母是否可以组成第一个字符串，每个字母只能用一次；两个字符串仅包含小写字母

自己的思路

将 ransom 转成 char 数组，遍历数组，然后在 magazines，进行对比，然后除去一个一样的字母

public static boolean canConstruct(String ransom, String magazine) {

	if (ransom.length() > magazine.length()) {
		return false;
	}

	char[] ransomChar = ransom.toCharArray();
	for (int i = 0; i < ransomChar.length; i++) {
		String str = String.valueOf(ransomChar[i]);
		if (magazine.indexOf(str) == -1) {
			return false;
		} else if (magazine.indexOf(str) == magazine.length() - 2) {
			magazine = magazine.substring(0, magazine.indexOf(str));
		} else {
			magazine = magazine.substring(0, magazine.indexOf(str)) + magazine.substring(magazine.indexOf(str) + 1);
		}
	}
	return true;
}

别人的算法

leetcode Ransom Note - 细语呢喃

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] cnt = new int[26];
        for (int i = 0; i < magazine.length(); i++) cnt[magazine.charAt(i) - 97]++;
		for (int i = 0; i < ransomNote.length(); i++) if (--cnt[ransomNote.charAt(i) - 97] < 0) return false;
		return true;
    }
}

我的理解
26 个元素的数组，每个元素存放字母的个数，如果需要的多于存在的 return false