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https://leetcode.com/problems/rotate-array/

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:

  • Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
  • Could you do it in-place with O(1) extra space?

解题思路:

O(nk)的解法

class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        while (k > 0) {
            int temp = nums[nums.length - 1];
            for (int i = nums.length - 1; i > 0; i--) {                
                nums[i] = nums[i - 1];                
            }
            nums[0] = temp;
            k--;
        }
    }
}

大神的解法,用reverse,O(n)

https://leetcode.com/problems/rotate-array/discuss/54250/Easy-to-read-Java-solution

public void rotate(int[] nums, int k) {
    k %= nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
}

public void reverse(int[] nums, int start, int end) {
    while (start < end) {
        int temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start++;
        end--;
    }
}

 

posted on 2018-12-25 00:36 NickyYe 阅读(...) 评论(...) 编辑 收藏