题目
【旋转数组】
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?
思路
k比数组长度大的情况，每nums.length次循环后数组还原，所以有效的k应该是k%nums.length。
那么，这时候问题转化为——“数组的后k位移动到数组的最前端”。
可以用翻转数组的方法来做：
[1,2,3,4,5,6,7]
先是后k位，以及前7-k位原地翻转：
[4,3,2,1]+[7,6,5]
此时再将整个数组进行一次翻转：
[5,6,7,1,2,3,4]
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 = k%nums.length;//k比数组长度大的情况，每nums.length次循环后数组还原，所以有效的k应该是k%nums.length
reverse(nums, nums.length-k, nums.length-1);
reverse(nums, 0, nums.length-k-1);
reverse(nums, 0, nums.length-1);
}
public void reverse(int[] nums, int start, int end){//用来反转数组中指定的一段
for( int temp; start < end ; start++,end--){
temp = nums[end];
nums[end] = nums[start];
nums[start] = temp;
}
}
}
提交结果
Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate Array.
Memory Usage: 38.2 MB, less than 30.38% of Java online submissions for Rotate Array.
