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‎README.md

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| [069][069-question] | [Sqrt(x)][069-tips] | [✅][069-java] | | [✅][069-kotlin] |
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| [070][070-question] | [Climbing Stairs][070-tips] | [✅][070-java] | | [✅][070-kotlin] |
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| [083][083-question] | [Remove Duplicates from Sorted List][083-tips] | [✅][083-java] | | [✅][083-kotlin] |
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| [088][088-question] | [Merge Sorted Array][088-tips] | [✅][088-java] | | |
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| [088][088-question] | [Merge Sorted Array][088-tips] | [✅][088-java] | | [✅][088-kotlin] |
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| [100][100-question] | [Same Tree][100-tips] | [✅][100-java] | | |
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| [101][101-question] | [Symmetric Tree][101-tips] | [✅][101-java] | | |
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| [104][104-question] | [Maximum Depth of Binary Tree][104-tips] | [✅][104-java] | | |
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| [674][674-question] | [Longest Continuous Increasing Subsequence][674-tips] | [✅][674-java] | [✅][674-js] | |
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| [680][680-question] | [Valid Palindrome II][680-tips] | [✅][680-java] | [✅][680-js] | |
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| [682][682-question] | [Baseball Game][682-tips] | [✅][682-java] | [✅][682-js] | |
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| [686][686-question] | [Repeated String Match][686-tips] | | [✅][686-js] | |
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| [686][686-question] | [Repeated String Match][686-tips] | [✅][686-java] | [✅][686-js] | |
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| [687][687-question] | [Longest Univalue Path][687-tips] | | [✅][687-js] | |
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| [693][693-question] | [Binary Number with Alternating Bits][693-tips] | | [✅][693-js] | |
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| [695][695-question] | [Max Area of Island][695-tips] | | [✅][695-js] | |
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[671-java]: ./src/_671/Solution.java
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[674-java]: ./src/_674/Solution.java
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[680-java]: ./src/_680/Solution.java
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[686-java]: ./src/_686/Solution.java
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[682-java]: ./src/_682/Solution.java
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[728-java]: ./src/_728/Solution.java
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[771-java]: ./src/_771/Solution.java
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[069-kotlin]: ./src/_069/kotlin/Solution.kt
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[070-kotlin]: ./src/_070/kotlin/Solution.kt
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[083-kotlin]: ./src/_083/kotlin/Solution.kt
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[088-kotlin]: ./src/_088/kotlin/Solution.kt
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[771-kotlin]: ./src/_771/kotlin/Solution.kt

‎src/_088/kotlin/Solution.kt

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package _088.kotlin
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import java.util.*
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/**
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* @author relish
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* @since 2018/04/24
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*/
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class Solution {
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fun merge(nums1: IntArray, m: Int, nums2: IntArray, n: Int): Unit {
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var index = n + m - 1
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var i = m - 1
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var j = n - 1
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while ((i >= 0 || j >= 0) && index >= 0) {
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if (j < 0 && i in 0..(m - 1)) {
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nums1[index--] = nums1[i--]
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continue
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}
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if (i < 0 && j in 0..(n - 1)) {
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nums1[index--] = nums2[j--]
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continue
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}
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if (nums1[i] > nums2[j]) {
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nums1[index--] = nums1[i--]
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} else {
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nums1[index--] = nums2[j--]
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}
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}
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}
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}
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fun main(args: Array<String>) {
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val arr1 = intArrayOf(1, 2, 3, 0, 0, 0)
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val arr2 = intArrayOf(2, 5, 6)
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Solution().merge(arr1, 3, arr2, 3)
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println(Arrays.toString(arr1))
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}

‎src/_686/Solution.java

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package _686;
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public class Solution {
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public int repeatedStringMatch(String A, String B) {
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if (A.length() > B.length()) {
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return A.contains(B) ? 1 : (A + A).contains(B) ? 2 : -1;
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}
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int aInB = B.indexOf(A);
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if (aInB >= A.length()) {
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return -1;
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}
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if (aInB == -1) {
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return (A + A).contains(B) ? 2 : -1;
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}
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int repeatCount = aInB == 0 ? 0 : 1;
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int iA = A.length() - aInB;
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int iB = 0;
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while (iB < B.length()) {
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if (iA == A.length()) {
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iA = 0;
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repeatCount++;
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}
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if (A.charAt(iA) != B.charAt(iB)) {
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return -1;
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}
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iA++;
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iB++;
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}
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return repeatCount;
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}
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public static void main(String[] args) {
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Solution solution = new Solution();
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System.out.println(solution.repeatedStringMatch("abcd", "cdabcdab"));
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System.out.println(solution.repeatedStringMatch("abcd", "ab"));
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System.out.println(solution.repeatedStringMatch("abcd", "da"));
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System.out.println(solution.repeatedStringMatch("a", "aa"));
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}
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}

‎tips/088/README.md

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题意是给两个已排序的数组 `nums1` 和 `nums2`，合并 `nums2` 到 `nums1` 中，两数组元素个数分别为 `m` 和 `n`，而且 `nums1` 数组的长度足够容纳 `m + n` 个元素，如果我们按顺序排下去，那肯定要开辟一个新数组来保存元素，如果我们选择逆序，这样利用 `nums1` 自身空间足矣，不会出现覆盖的情况，依次把大的元素插入到 `nums1` 的末尾，确保 `nums2` 中的元素全部插入到 `nums1` 即可。
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Java:
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```java
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class Solution {
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public void merge(int[] nums1, int m, int[] nums2, int n) {
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}
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```
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kotlin(208ms/100.00%)
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```kotlin
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class Solution {
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fun merge(nums1: IntArray, m: Int, nums2: IntArray, n: Int): Unit {
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var index = n + m - 1
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var i = m - 1
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var j = n - 1
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while ((i >= 0 || j >= 0) && index >= 0) {
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if (j < 0 && i in 0..(m - 1)) {
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nums1[index--] = nums1[i--]
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continue
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}
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if (i < 0 && j in 0..(n - 1)) {
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nums1[index--] = nums2[j--]
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continue
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}
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if (nums1[i] > nums2[j]) {
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nums1[index--] = nums1[i--]
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} else {
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nums1[index--] = nums2[j--]
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}
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}
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}
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}
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```
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## 结语
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‎tips/686/README.md

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[xxxx][title]
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[Repeated String Match][title]
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## Description
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// 抄题目
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Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.
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**Example:**
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For example, with A = "abcd" and B = "cdabcdab".
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```
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// 抄Example
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```
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Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").
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**Note:**
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// Note
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**Tags:** // tags
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The length of `A` and `B` will be between 1 and 10000.
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**Tags:** [String](https://leetcode.com/tag/string/)
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## 思路 1
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// 贴一些关键代码,说一些解题思路
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// (同一种语言可以写多种思路，与某种语言思路相同的另一种语言的思路无须赘述，但可以把代码贴在后面)
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```java
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## 思路
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24-
```
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```javascript
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判断重复的A字符串能否使B字符串成为子字符串，若能，计算出至少需要重复几次。使用A字符串对B字符串重复进行匹配即可。但在匹配之前可通过一些已知条件减少一些计算：
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```
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- 如果A的长度大于B，假如重复A字符串能使B成为子字符串，那么至多只需重复两次。因为如果重复三次才行，那么B的长度肯定大于A，就矛盾了。
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- 如果A的长度小于B，用`IndexOf()`方法计算字符串A在B中首次匹配的位置，记作`aInB`。
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- 如果`aInB`>=`A.length()`，则说明在B的前面一段（比A长或相等）字符串中，没有和A匹配的子字符串，这跟连续的A字符串这个要求不符，因此可以排除。
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- 如果`aInB`==`-1`，则B字符串只有可能是两个连续A字符串的子字符串。
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- 接着循环用A字符串对B字符串进行匹配。
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## 思路 2
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// 贴一些关键代码,说一些解题思路
31-
```java
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33-
```
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## 思路 3
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// 贴一些关键代码,说一些解题思路
37-
```kotlin
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**Java:**
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29+
```java
30+
public int repeatedStringMatch(String A, String B) {
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if (A.length() > B.length()) {
32+
return A.contains(B) ? 1 : (A + A).contains(B) ? 2 : -1;
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}
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int aInB = B.indexOf(A);
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if (aInB >= A.length()) {
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return -1;
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}
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if (aInB == -1) {
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return (A + A).contains(B) ? 2 : -1;
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}
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int repeatCount = aInB == 0 ? 0 : 1;
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int iA = A.length() - aInB;
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int iB = 0;
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while (iB < B.length()) {
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if (iA == A.length()) {
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iA = 0;
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repeatCount++;
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}
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if (A.charAt(iA) != B.charAt(iB)) {
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return -1;
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}
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iA++;
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iB++;
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}
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return repeatCount;
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}
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```
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## 结语
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如果你同我们一样热爱数据结构、算法、LeetCode，可以关注我们 GitHub 上的 LeetCode 题解：[LeetCode-Solution][ls]
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[title]: https://leetcode.com/problems/xxxx
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[title]: https://leetcode.com/problems/repeated-string-match/description/
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[ls]: https://github.com/RichCodersAndMe/LeetCode-Solution

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