[leetcode]189.Rotate Array

题目

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]

解法一

思路

比较暴力无脑的解法，直接先将数组后面k个元素先移动到其他数组，然后将前面nums.length-k个元素向后移动k位，然后再将其他数组的k个元素移回数组前k位即可。时间复杂度O(n)，空间复杂度也是O(n)。

代码

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

解法二

思路

先将数组整体逆转一下，然后将前k位逆转一下，再将后nums.length-k位逆转一下。时间复杂度O(n)，空间复杂度O(1)。

代码

class 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);
    }
    private void reverse(int[] nums, int left, int right) {
        while(left < right) {
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}