You are given an integer array nums of length n. Please construct an answer array ans with a length of 2n, the array subscript starts counting from 0, and for all i where 0 <= i < n, satisfy all the following requirements:
- ans[i] == nums[i]
- ans[i + n] == nums[i]
Specifically, ans is formed by concatenating two nums arrays.
Returns the array ans.
Example 1
输入：nums = [1,2,1]
输出：[1,2,1,1,2,1]
解释：数组 ans 按下述方式形成：
- ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]]
- ans = [1,2,1,1,2,1]
Example 2
输入：nums = [1,3,2,1]
输出：[1,3,2,1,1,3,2,1]
解释：数组 ans 按下述方式形成：
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]
The first method
uses a loop assignment
class Concatenation {
public int[] getConcatenation(int[] nums) {
int[] ans = new int[nums.length * 2];
for (int i = 0; i < nums.length; i++) {
//数组复制
ans[i] = nums[i];
ans[i + nums.length] = nums[i];
}
return ans;
}
}
The second
way to use the array comes with
class Concatenation2
{
public int[] getConcatenation(int[] nums) {
int n = nums.length;
int[] ans = Arrays.copyOf(nums, n*2);
System.arraycopy(nums, 0, ans, n, n);
return ans;
}
}