Leetcode学习笔记：#771 Jewels and Stones

原创于 2019-05-06 17:01:46 发布 · 101 阅读

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此博客是Leetcode #771 Jewels and Stones的学习笔记。题目给出代表宝石类型的字符串J和代表拥有石头的字符串S，要计算拥有的石头中有多少是宝石。实现思路是两次遍历，第一次将S的字符存入charToFreq并计数，第二次将相同字符对应的值累加。

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Leetcode学习笔记：#771 Jewels and Stones

You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so “a” is considered a different type of stone from “A”.

实现：

public int numJewlsInStones(String J, String S){
	if(J.length() == 0 || S.length() == 0){
		return 0;
	}

	int[] charToFreq = new int[256];
	for(char ch:S.toCharArray()){
		charToFreq[ch]++;
	}
	int numJewels = 0;
	for(char ch:J.toCharArray()){
		if(charToFreq[ch] > 0){
			numJewels += charToFreq[ch];
		}
	}
	return numJewels;
}

思路：
两次遍历。第一次遍历把S的每个char遍历进charToFreq，按ASCII码分配，每有一个相同的char便加1，第二次遍历每发现一个相同的char，便把charToFreq中该char相对应的value加入numJewels中。