1967. 作为子字符串出现在单词中的字符串数目
1967. 作为子字符串出现在单词中的字符串数目
题目
Given an array of strings patterns
and a string word
, return _thenumber of strings in _patterns
_that exist as asubstring in _word
.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: patterns = ["a","abc","bc","d"], word = "abc"
Output: 3
Explanation:
- "a" appears as a substring in "a bc".
- "abc" appears as a substring in "abc ".
- "bc" appears as a substring in "a bc ".
- "d" does not appear as a substring in "abc".
3 of the strings in patterns appear as a substring in word.
Example 2:
Input: patterns = ["a","b","c"], word = "aaaaabbbbb"
Output: 2
Explanation:
- "a" appears as a substring in "a a aaabbbbb".
- "b" appears as a substring in "aaaaabbbb b ".
- "c" does not appear as a substring in "aaaaabbbbb".
2 of the strings in patterns appear as a substring in word.
Example 3:
Input: patterns = ["a","a","a"], word = "ab"
Output: 3
Explanation: Each of the patterns appears as a substring in word "a b".
Constraints:
1 <= patterns.length <= 100
1 <= patterns[i].length <= 100
1 <= word.length <= 100
patterns[i]
andword
consist of lowercase English letters.
题目大意
给你一个字符串数组 patterns
和一个字符串 word
,统计 patterns
中有多少个字符串是 word
的子字符串。返回字符串数目。
子字符串 是字符串中的一个连续字符序列。
示例 1:
输入: patterns = ["a","abc","bc","d"], word = "abc"
输出: 3
解释:
- "a" 是 "a bc" 的子字符串。
- "abc" 是 "abc " 的子字符串。
- "bc" 是 "a bc " 的子字符串。
- "d" 不是 "abc" 的子字符串。
patterns 中有 3 个字符串作为子字符串出现在 word 中。
示例 2:
输入: patterns = ["a","b","c"], word = "aaaaabbbbb"
输出: 2
解释:
- "a" 是 "a a aaabbbbb" 的子字符串。
- "b" 是 "aaaaabbbb b " 的子字符串。
- "c" 不是 "aaaaabbbbb" 的字符串。
patterns 中有 2 个字符串作为子字符串出现在 word 中。
示例 3:
输入: patterns = ["a","a","a"], word = "ab"
输出: 3
解释: patterns 中的每个字符串都作为子字符串出现在 word "a b" 中。
提示:
1 <= patterns.length <= 100
1 <= patterns[i].length <= 100
1 <= word.length <= 100
patterns[i]
和word
由小写英文字母组成
解题思路
- 初始化计数器
count
为 0,用于记录符合条件的子串数量。 - 遍历
patterns
数组中的每个模式pattern
:- 如果
word
包含pattern
,将count
加 1。
- 如果
- 遍历结束后,返回
count
。
复杂度分析
时间复杂度:
O(m * n)
- 遍历
patterns
的时间复杂度是O(m)
,其中m
是patterns
的长度。 - 对每个
pattern
调用.includes()
方法的时间复杂度为O(n)
,其中n
是word
的长度。 - 总体时间复杂度为
O(m * n)
。
- 遍历
空间复杂度:
O(1)
,使用了常量空间。
代码
/**
* @param {string[]} patterns
* @param {string} word
* @return {number}
*/
var numOfStrings = function (patterns, word) {
let count = 0;
for (let pattern of patterns) {
if (word.includes(pattern)) {
count++;
}
}
return count;
};