添加数组右移k次的算法

董林 · 董林 · commit 8544f50dadc0 · 2018-07-12T14:02:08.000+08:00
diff --git a/src/main/java/com/crossoverjie/algorithm/ArrayKShift.java b/src/main/java/com/crossoverjie/algorithm/ArrayKShift.java
@@ -0,0 +1,75 @@
+package com.crossoverjie.algorithm;
+
+import java.util.Arrays;
+
+/**
+ * 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>
+ *
+ * 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
+ * 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
+ * 2 同理后 len-k个元素只需要翻转就完成数组的k次移动
+ *
+ * @author 656369960@qq.com
+ * @date 12/7/2018 1:38 PM
+ * @since 1.0
+ */
+public class ArrayKShift {
+
+    public void arrayKShift(int[] array, int k) {
+
+        /**
+         * constrictions
+         */
+
+        if (array == null || 0 == array.length) {
+            return ;
+        }
+
+        k = k % array.length;
+
+        if (0 > k) {
+            return;
+        }
+
+
+        /**
+         * reverse array , e.g: [1, 2, 3 ,4] to [4,3,2,1]
+         */
+
+        for (int i = 0; i < array.length / 2; i++) {
+            int tmp = array[i];
+            array[i] = array[array.length - 1 - i];
+            array[array.length - 1 - i] = tmp;
+        }
+
+        /**
+         * first k element reverse
+         */
+        for (int i = 0; i < k / 2; i++) {
+            int tmp = array[i];
+            array[i] = array[k - 1 - i];
+            array[k - 1 - i] = tmp;
+        }
+
+        /**
+         * last length - k element reverse
+         */
+
+        for (int i = k; i < k + (array.length - k ) / 2; i ++) {
+            int tmp = array[i];
+            array[i] = array[array.length - 1 - i + k];
+            array[array.length - 1 - i + k] = tmp;
+        }
+    }
+
+    public static void main(String[] args) {
+        int[] array = {1, 2, 3 ,4, 5, 6, 7};
+        ArrayKShift shift = new ArrayKShift();
+        shift.arrayKShift(array, 6);
+
+        Arrays.stream(array).forEach(o -> {
+            System.out.println(o);
+        });
+
+    }
+}
