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本帖最后由 Seawolf 于 2020-9-16 04:15 编辑 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?
class Solution {
public void rotate(int[] nums, int k) {
if(nums.length ==0 || nums.length ==1) return;
if(k == 0) return;
if(k > nums.length) k = k % nums.length;
int[] res = new int[nums.length];
for(int i = 0; i < nums.length; i++){
if(i < k){
res[i] = nums[i+nums.length-k];
}else if(i >= k){
res[i] = nums[i-k];
}
}
for(int i = 0; i < nums.length; i++){
nums[i] = res[i];
}
}
}
class Solution {
public void rotate(int[] nums, int k) {
if(nums.length == 0 || nums.length == 1) return;
if(k>nums.length) k = k % nums.length;
reverse(nums,0, nums.length-1);
reverse(nums,0,k-1);
reverse(nums,k,nums.length-1);
}
public void reverse(int[] nums, int start, int end){
while(end - start >= 1 && start < end){
int i = nums[start];
nums[start] = nums[end];
nums[end] =i;
start++;
end--;
}
}
}
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