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69 lines
2.1 KiB
Markdown
69 lines
2.1 KiB
Markdown
# [1668. Maximum Repeating Substring](https://leetcode.com/problems/maximum-repeating-substring/)
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## 题目
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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`.
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Given strings `sequence` and `word`, return *the **maximum `k`-repeating value** of `word` in `sequence`*.
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**Example 1:**
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```
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Input: sequence = "ababc", word = "ab"
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Output: 2
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Explanation: "abab" is a substring in "ababc".
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```
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**Example 2:**
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```
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Input: sequence = "ababc", word = "ba"
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Output: 1
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Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".
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```
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**Example 3:**
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```
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Input: sequence = "ababc", word = "ac"
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Output: 0
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Explanation: "ac" is not a substring in "ababc".
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```
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**Constraints:**
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- `1 <= sequence.length <= 100`
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- `1 <= word.length <= 100`
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- `sequence` and `word` contains only lowercase English letters.
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## 题目大意
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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 。
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## 解题思路
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- 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 。
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## 代码
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```go
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package leetcode
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import (
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"strings"
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)
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func maxRepeating(sequence string, word string) int {
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for i := len(sequence) / len(word); i >= 0; i-- {
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tmp := ""
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for j := 0; j < i; j++ {
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tmp += word
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}
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if strings.Contains(sequence, tmp) {
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return i
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}
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}
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return 0
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}
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``` |