转载(mark)
https://blog.by24.cn/archives/leetcode-count-the-repetitions.html
func getMaxRepetitions(s1 string, n1 int, s2 string, n2 int) int {len1, len2 := len(s1), len(s2)index1, index2 := 0, 0 // 注意此处直接使用 Ra Rb 的下标,不取模if len1 == 0 || len2 == 0 || len1*n1 < len2*n2 {return 0}map1, map2 := make(map[int]int), make(map[int]int)ans := 0 // 注意,此处存储的是 Ra 中 Sb 的个数,而非 Ra 中 Rb 的个数for index1/len1 < n1 { // 遍历整个 Raif index1%len1 == len1-1 { //在 Sa 末尾if val, ok := map1[index2%len2]; ok { // 出现了循环,进行快进cycleLen := index1/len1 - val/len1 // 每个循环占多少个 SacycleNum := (n1 - 1 - index1/len1) / cycleLen // 还有多少个循环cycleS2Num := index2/len2 - map2[index2%len2]/len2 // 每个循环含有多少个 Sbindex1 += cycleNum * cycleLen * len1 // 将 Ra 快进到相应的位置ans += cycleNum * cycleS2Num // 把快进部分的答案数量加上} else { // 第一次,注意存储的是未取模的map1[index2%len2] = index1map2[index2%len2] = index2}}if s1[index1%len1] == s2[index2%len2] {if index2%len2 == len2-1 {ans += 1}index2 += 1}index1 += 1}return ans / n2
}