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/**
* <p>给你一个数组,将数组中的元素向右轮转 <code>k</code><em> </em>个位置,其中 <code>k</code><em> </em>是非负数。</p>
*
* <p> </p>
*
* <p><strong>示例 1:</strong></p>
*
* <pre>
* <strong>输入:</strong> nums = [1,2,3,4,5,6,7], k = 3
* <strong>输出:</strong> <code>[5,6,7,1,2,3,4]</code>
* <strong>解释:</strong>
* 向右轮转 1 步: <code>[7,1,2,3,4,5,6]</code>
* 向右轮转 2 步: <code>[6,7,1,2,3,4,5]
* </code>向右轮转 3 步: <code>[5,6,7,1,2,3,4]</code>
* </pre>
*
* <p><strong>示例 2:</strong></p>
*
* <pre>
* <strong>输入:</strong>nums = [-1,-100,3,99], k = 2
* <strong>输出:</strong>[3,99,-1,-100]
* <strong>解释:</strong>
* 向右轮转 1 步: [99,-1,-100,3]
* 向右轮转 2 步: [3,99,-1,-100]</pre>
*
* <p> </p>
*
* <p><strong>提示:</strong></p>
*
* <ul>
* <li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
* <li><code>-2<sup>31</sup> <= nums[i] <= 2<sup>31</sup> - 1</code></li>
* <li><code>0 <= k <= 10<sup>5</sup></code></li>
* </ul>
*
* <p> </p>
*
* <p><strong>进阶:</strong></p>
*
* <ul>
* <li>尽可能想出更多的解决方案,至少有 <strong>三种</strong> 不同的方法可以解决这个问题。</li>
* <li>你可以使用空间复杂度为 <code>O(1)</code> 的 <strong>原地 </strong>算法解决这个问题吗?</li>
* </ul>
*
* <ul>
* </ul>
*
* <ul>
* </ul>
* <div><div>Related Topics</div><div><li>数组</li><li>数学</li><li>双指针</li></div></div><br><div><li>👍 1190</li><li>👎 0</li></div>
*/
package leetcode1;
public class RotateArray {
public static void main(String[] args) {
Solution solution = new RotateArray().new Solution();
}
/**
* 数组多次翻转
*/
class Solution {
public void rotate(int[] nums, int k) {
k = k % nums.length;
solve(nums, 0, nums.length - 1);
solve(nums, 0, k - 1);
solve(nums, k, nums.length - 1);
}
private void solve(int[] nums, int i, int j) {
while (i < j) {
swap(nums, i++, j--);
}
}
private void swap(int[] nums, int i, int j) {
int k = nums[i];
nums[i] = nums[j];
nums[j] = k;
}
}
/**
* 使用额外的数组
*/
class Solution2 {
public void rotate(int[] nums, int k) {
int[] res = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
res[(i + k) % nums.length] = nums[i];
}
System.arraycopy(res, 0, nums, 0, nums.length);
}
}
}