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_189.java
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package com.fishercoder.solutions;
import java.util.ArrayList;
import java.util.List;
/**
* 189. 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?
* */
public class _189 {
public static class Solution1 {
/**
* O(n) space
* O(n) time
* */
public void rotate(int[] nums, int k) {
int len = nums.length;
int[] tmp = new int[len];
for (int i = 0; i < len; i++) {
tmp[(i + k) % len] = nums[i];
}
for (int i = 0; i < len; i++) {
nums[i] = tmp[i];
}
}
}
public static class Solution2 {
/**
* O(1) space
* O(n) time
* */
public void rotate(int[] nums, int k) {
int tmp;
for (int i = 0; i < k; i++) {
tmp = nums[nums.length - 1];
for (int j = nums.length - 1; j > 0; j--) {
nums[j] = nums[j - 1];
}
nums[0] = tmp;
}
}
}
public static class Solution3 {
/**
* My original idea and got AC'ed.
* One thing to notice is that when k > nums.length, we'll continue to rotate the array, it just becomes k -= nums.length
*/
public static void rotate(int[] nums, int k) {
if (k == 0 || k == nums.length) {
return;
}
if (k > nums.length) {
k -= nums.length;
}
List<Integer> tmp = new ArrayList();
int i = 0;
if (nums.length - k >= 0) {
i = nums.length - k;
for (; i < nums.length; i++) {
tmp.add(nums[i]);
}
} else {
i = nums.length - 1;
for (; i >= 0; i--) {
tmp.add(nums[i]);
}
}
for (i = 0; i < nums.length - k; i++) {
tmp.add(nums[i]);
}
for (i = 0; i < tmp.size(); i++) {
nums[i] = tmp.get(i);
}
}
}
}