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189. Rotate Array
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/**********
//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?
**********/
/********** Solution1 **********/
class Solution {
public void rotate(int[] nums, int k) {
k = (nums.length + (k % nums.length)) % nums.length;
int temp;
for (int i = 0, j = nums.length - 1; i < j; i++, j--) {
temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
for (int i = 0, j = k-1; i < j; i++, j--) {
temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
for (int i = k, j = nums.length - 1; i < j; i++, j--) {
temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
}
// [1,2,3,4,5,6,7] and k = 3
// [7,6,5,4,3,2,1]
// [5,6,7,4,3,2,1]
// [5,6,7,1,2,3,4]
/********** Solution1 **********/
/********** Solution2 **********/
class Solution {
public void rotate(int[] nums, int k) {
if (nums == null || k < 0) {
return;
}
int temp;
for (int i = 1; i <= k; i++) {
for (int j = nums.length - 1; j > 0; j--) {
temp = nums[j];
nums[j] = nums[j-1];
nums[j-1] = temp;
}
}
}
}
// It's like bubble rotate
// i=0
// 0 1 2 3 4 5 6
// 0 1 2 3 4 6 5
// ...
// 6 0 1 2 3 4 5
// i=1
// 6 0 1 2 3 5 4
// ...
// 5 6 0 1 2 3 4
// i=2
// 5 6 0 1 2 4 3
// ...
// 4 5 6 0 1 2 3
/********** Solution2 **********/