題目

https://leetcode.com/problems/repeated-string-match/

題解

套了 KMP 模板，O(n) 復雜度。分析如下。

class Solution {public int repeatedStringMatch(String a, String b) {int repeatedTimes = Math.max((int) Math.ceil((float) 2 * b.length() / a.length()), 2);int index = indexOf(a.repeat(repeatedTimes), b);if (index == -1) return -1;return (int) Math.ceil((float) (index + b.length()) / a.length());}// KMP, find index of s2 from s1public static int indexOf(String s1, String s2) {if (s1 == null || s2 == null || s2.length() < 1 || s1.length() < s2.length()) {return -1;}char[] str1 = s1.toCharArray();char[] str2 = s2.toCharArray();int x = 0;int y = 0;// O(M) m <= nint[] next = nextArray(str2);// O(N)while (x < str1.length && y < str2.length) {if (str1[x] == str2[y]) {x++;y++;} else if (next[y] == -1) { // y == 0x++;} else {y = next[y];}}return y == str2.length ? x - y : -1;}public static int[] nextArray(char[] str2) {if (str2.length == 1) {return new int[]{-1};}int[] next = new int[str2.length];next[0] = -1;next[1] = 0;int i = 2; // 目前在哪個位置上求next數組的值int cn = 0; // 當前是哪個位置的值再和i-1位置的字符比較while (i < next.length) {if (str2[i - 1] == str2[cn]) { // 配成功的時候next[i++] = ++cn;} else if (cn > 0) {cn = next[cn];} else {next[i++] = 0;}}return next;} }

總結

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