README_EN.md

1304. Find N Unique Integers Sum up to Zero

中文文档

Description

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

 

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

Solutions

Python3

class Solution:
    def sumZero(self, n: int) -> List[int]:
        ans = []
        for i in range(n >> 1):
            ans.append(i + 1)
            ans.append(-(i + 1))
        if n & 1:
            ans.append(0)
        return ans

class Solution:
    def sumZero(self, n: int) -> List[int]:
        ans = list(range(1, n))
        ans.append(-sum(ans))
        return ans

Java

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

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

C++

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

class Solution {
public:
    vector<int> sumZero(int n) {
        vector<int> ans(n);
        iota(ans.begin(), ans.end(), 1);
        ans[n - 1] = -(n - 1) * n / 2;
        return ans;
    }
};

Go

func sumZero(n int) []int {
	ans := make([]int, n)
	for i, j := 1, 0; i <= n/2; i, j = i+1, j+1 {
		ans[j] = i
		j++
		ans[j] = -i
	}
	return ans
}

func sumZero(n int) []int {
	ans := make([]int, n)
	for i := 1; i < n; i++ {
		ans[i] = i
	}
	ans[0] = -n * (n - 1) / 2
	return ans
}

TypeScript

function sumZero(n: number): number[] {
    const ans = new Array(n).fill(0);
    for (let i = 1, j = 0; i <= n / 2; ++i) {
        ans[j++] = i;
        ans[j++] = -i;
    }
    return ans;
}

function sumZero(n: number): number[] {
    const ans = new Array(n).fill(0);
    for (let i = 1; i < n; ++i) {
        ans[i] = i;
    }
    ans[0] = -((n * (n - 1)) / 2);
    return ans;
}

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