QUESTION:
Given the array nums consisting of 2n elements in the form [x1,x2,…,xn,y1,y2,…,yn].
Return the array in the form [x1,y1,x2,y2,…,xn,yn].
Example 1:
Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7]
Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
Example 2:
Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]
Example 3:
Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]
Constraints:
1 <= n <= 500
nums.length == 2n
1 <= nums[i] <= 10^3
EXPLANATION:
简单的题目,使用双指针,并用idx来判断奇偶。
思路:
- 创建双指针,一个用来指向奇数,一个用来指向偶数。
- 创建一个长度为2n的数组,用来摆放结果,同时创建一个idx用来指向此时应该摆放的位置。
- 如果当前的idx能够被2整除,那么说明是x的,我们就将odd的值填入,同时将两者自增1.
- 如果当前的idx是不能被2整除,那么说明是y的,我们就将even的值添加,同时将两者自增1.
SOLUTION:
class Solution {
public int[] shuffle(int[] nums, int n) {
int[] result = new int[2*n];
int idx = 0;
int odd = 0;
int even = n;
while (idx<nums.length){
if(idx % 2 == 0)
result[idx++] = nums[odd++];
else
result[idx++] = nums[even++];
}
return result;
}
}