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LeetCode 189. Rotate Array (旋轉數組)

note div 題目 方法 leetcode array beats not htm

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         int
turns = 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)