1668. Maximum Repeating Substring

题目

For a string sequence, a string word is k-repeating if word concatenated k times is a substring of sequence. The word's maximum k-repeating value is the highest value k where word is k-repeating in sequence. If word is not a substring of sequence, word's maximum k-repeating value is 0.

Given strings sequence and word, return the maximum k-repeating value of word in sequence.

Example 1:

Input: sequence = "ababc", word = "ab"
Output: 2
Explanation: "abab" is a substring in "ababc".

Example 2:

Input: sequence = "ababc", word = "ba"
Output: 1
Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".

Example 3:

Input: sequence = "ababc", word = "ac"
Output: 0
Explanation: "ac" is not a substring in "ababc".

Constraints:

1 <= sequence.length <= 100
1 <= word.length <= 100
sequence and word contains only lowercase English letters.

题目大意

给你一个字符串 sequence ，如果字符串 word 连续重复 k 次形成的字符串是 sequence 的一个子字符串，那么单词 word 的重复值为 k 。单词 word 的最大重复值是单词 word 在 sequence 中最大的重复值。如果 word 不是 sequence 的子串，那么重复值 k 为 0 。给你一个字符串 sequence 和 word ，请你返回最大重复值 k 。

解题思路

循环叠加构造 word，每次构造出新的 word 都在 sequence 查找一次，如果找到就输出叠加次数，否则继续叠加构造，直到字符串长度和 sequence 一样长，最终都没有找到则输出 0 。

代码

package leetcode

import (
	"strings"
)

func maxRepeating(sequence string, word string) int {
	for i := len(sequence) / len(word); i >= 0; i-- {
		tmp := ""
		for j := 0; j < i; j++ {
			tmp += word
		}
		if strings.Contains(sequence, tmp) {
			return i
		}
	}
	return 0
}

Uh oh!

pFad - (p)hone/(F)rame/(a)nonymizer/(d)eclutterfier! Saves Data!