1、题干
给你一个由不同字符组成的字符串 allowed
和一个字符串数组 words
。如果一个字符串的每一个字符都在 allowed
中,就称这个字符串是 一致字符串 。
请你返回 words
数组中 一致字符串 的数目。
示例 1:
输入:allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
输出:2
解释:字符串 "aaab" 和 "baa" 都是一致字符串,因为它们只包含字符 'a' 和 'b' 。
示例 2:
输入:allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
输出:7
解释:所有字符串都是一致的。
示例 3:
输入:allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
输出:4
解释:字符串 "cc","acd","ac" 和 "d" 是一致字符串。
提示:
1 <= words.length <= 104
1 <= allowed.length <= 26
1 <= words[i].length <= 10
allowed
中的字符 互不相同 。words[i]
和allowed
只包含小写英文字母。
Problem: 1684. 统计一致字符串的数目
[TOC]
思路
reduce 跟 for of 耗费的时间空间差距还挺大
Code - for of
function countConsistentStrings(allowed: string, words: string[]): number {
const dict = new Set(allowed);
let res = 0;
loop: for (const s of words) {
for (const c of s) {
if (!dict.has(c)) continue loop;
}
res++;
}
return res;
};
Code - reduce
function countConsistentStrings(allowed: string, words: string[]): number {
const dict = new Set(allowed);
return words.reduce((a, s) => a + +([].every.call(s, (c) => dict.has(c))), 0);
};