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Merge pull request crossoverJie#63 from ryandonglin/master
Close issue 49
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README.md

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@@ -82,6 +82,7 @@ Java 知识点,继续完善中。
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- [两个栈实现队列](https://github.com/crossoverJie/Java-Interview/blob/master/src/main/java/com/crossoverjie/algorithm/TwoStackQueue.java)
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- [动手实现一个 LRU cache](http://crossoverjie.top/2018/04/07/algorithm/LRU-cache/)
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- [链表排序](./src/main/java/com/crossoverjie/algorithm/LinkedListMergeSort.java)
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- [数组右移k次](./src/main/java/com/crossoverjie/algorithm/ArrayKShift.java)
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### Netty 相关
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- [SpringBoot 整合长连接心跳机制](https://crossoverjie.top/2018/05/24/netty/Netty(1)TCP-Heartbeat/)
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package com.crossoverjie.algorithm;
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import java.util.Arrays;
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/**
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* 数组右移K次, 原数组<code> [1, 2, 3, 4, 5, 6, 7]</code> 右移3次后结果为 <code>[5,6,7,1,2,3,4]</code>
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*
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* 基本思路:不开辟新的数组空间的情况下考虑在原属组上进行操作
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* 1 将数组倒置,这样后k个元素就跑到了数组的前面,然后反转一下即可
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* 2 同理后 len-k个元素只需要翻转就完成数组的k次移动
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*
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* @date 12/7/2018 1:38 PM
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* @since 1.0
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*/
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public class ArrayKShift {
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public void arrayKShift(int[] array, int k) {
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/**
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* constrictions
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*/
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if (array == null || 0 == array.length) {
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return ;
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}
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k = k % array.length;
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if (0 > k) {
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return;
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}
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/**
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* reverse array , e.g: [1, 2, 3 ,4] to [4,3,2,1]
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*/
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for (int i = 0; i < array.length / 2; i++) {
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int tmp = array[i];
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array[i] = array[array.length - 1 - i];
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array[array.length - 1 - i] = tmp;
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}
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/**
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* first k element reverse
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*/
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for (int i = 0; i < k / 2; i++) {
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int tmp = array[i];
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array[i] = array[k - 1 - i];
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array[k - 1 - i] = tmp;
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}
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/**
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* last length - k element reverse
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*/
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for (int i = k; i < k + (array.length - k ) / 2; i ++) {
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int tmp = array[i];
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array[i] = array[array.length - 1 - i + k];
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array[array.length - 1 - i + k] = tmp;
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}
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}
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public static void main(String[] args) {
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int[] array = {1, 2, 3 ,4, 5, 6, 7};
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ArrayKShift shift = new ArrayKShift();
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shift.arrayKShift(array, 6);
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Arrays.stream(array).forEach(o -> {
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System.out.println(o);
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});
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}
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}

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