1304 Find N Unique Integers Sum up to Zero 和為零的N個(gè)唯一整數(shù)

Description:
Given an integer n, return any array containing n unique integers such that they add up to 0.

Example:

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

1 <= n <= 1000

題目描述:
給你一個(gè)整數(shù) n，請你返回 任意 一個(gè)由 n 個(gè) 各不相同 的整數(shù)組成的數(shù)組，并且這 n 個(gè)數(shù)相加和為 0 。

示例 :

示例 1：

輸入：n = 5
輸出：[-7,-1,1,3,4]
解釋：這些數(shù)組也是正確的 [-5,-1,1,2,3]，[-3,-1,2,-2,4]。

示例 2：

輸入：n = 3
輸出：[-1,0,1]

示例 3：

輸入：n = 1
輸出：[0]

提示：

1 <= n <= 1000

思路:

從 1 - n到 n - 1按照步長為 2加入到結(jié)果數(shù)組即可
時(shí)間復(fù)雜度O(n), 空間復(fù)雜度O(1)

代碼:
C++:

class Solution 
{
public:
    vector<int> sumZero(int n) 
    {
        vector<int> result(n);
        for (int num = 1 - n, i = 0; num < n; num += 2, i++) result[i] = num;
        return result;
    }
};

Java:

class Solution {
    public int[] sumZero(int n) {
        int result[] = new int[n];
        for (int i = 0, num = 1 - n; i < n; i++, num += 2) result[i] = num;
        return result;
    }
}

Python:

class Solution:
    def sumZero(self, n: int) -> List[int]:
        return range(1 - n, n, 2)