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1304. Find N Unique Integers Sum up to Zero
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
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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: 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: 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 -
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 } -
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; }