LeetCode 189. Rotate Array (旋轉數組)
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7]
is rotated to [5,6,7,1,2,3,4]
.
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
[show hint]
Hint:Could you do it in-place with O(1) extra space?
Related problem: Reverse Words in a String II
題目標簽:Array
題目給了我們一個數組 和 k。 讓我們 旋轉數組 k 次。
這裏有一個很巧妙的方法:
利用數組的length - k 把數組 分為兩半;
reverse 左邊和右邊的數組;
reverse 總數組。
舉一個例子:
1 2 3 4 5 6 7 如果k = 3 的話, 會變成 5 6 7 1 2 3 4
1 2 3 4 5 6 7 middle = 7 - 3 = 4,分為左邊 4個數字,右邊 3個數字
4 3 2 1 7 6 5 分別把左右reverse 一下
5 6 7 1 2 3 4 把總數組reverse 一下就會得到答案
Java Solution:
Runtime beats 15.37%
完成日期:04/13/2017
關鍵詞:Array
關鍵點:利用k把數組分兩半;reverse左右兩邊數組;reverse總數組
1 public class Solution 2 { 3 public void rotate(int[] nums, int k) 4 { 5 if(nums == null || nums.length == 0 || k % nums.length == 0) 6 return; 7 8 intturns = k % nums.length; 9 int middle = nums.length - turns; 10 11 reverse(nums, 0, middle-1); // reverse left part 12 reverse(nums, middle, nums.length-1); // reverse right part 13 reverse(nums, 0, nums.length-1); // reverse whole part 14 } 15 16 public void reverse(int[] arr, int s, int e) 17 { 18 while(s < e) 19 { 20 int temp = arr[s]; 21 arr[s] = arr[e]; 22 arr[e] = temp; 23 24 s++; 25 e--; 26 } 27 } 28 }
參考資料:
http://www.cnblogs.com/grandyang/p/4298711.html
LeetCode 算法題目列表 - LeetCode Algorithms Questions List
LeetCode 189. Rotate Array (旋轉數組)