README_EN.md

2239. Find Closest Number to Zero

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Description

Given an integer array nums of size n, return the number with the value closest to 0 in nums. If there are multiple answers, return the number with the largest value.

 

Example 1:

Input: nums = [-4,-2,1,4,8]
Output: 1
Explanation:
The distance from -4 to 0 is |-4| = 4.
The distance from -2 to 0 is |-2| = 2.
The distance from 1 to 0 is |1| = 1.
The distance from 4 to 0 is |4| = 4.
The distance from 8 to 0 is |8| = 8.
Thus, the closest number to 0 in the array is 1.

Example 2:

Input: nums = [2,-1,1]
Output: 1
Explanation: 1 and -1 are both the closest numbers to 0, so 1 being larger is returned.

 

Constraints:

• 1 <= n <= 1000
• -105 <= nums[i] <= 105

Solutions

Python3

class Solution:
    def findClosestNumber(self, nums: List[int]) -> int:
        ans, d = 0, 1000000
        for v in nums:
            if (t := abs(v)) < d or (t == d and v > ans):
                ans, d = v, t
        return ans

Java

class Solution {
    public int findClosestNumber(int[] nums) {
        int ans = 0, d = 1000000;
        for (int v : nums) {
            int t = Math.abs(v);
            if (t < d || (t == d && v > ans)) {
                ans = v;
                d = t;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findClosestNumber(vector<int>& nums) {
        int ans = 0, d = 1e6;
        for (int& v : nums) {
            int t = abs(v);
            if (t < d || (t == d && v > ans)) {
                ans = v;
                d = t;
            }
        }
        return ans;
    }
};

Go

func findClosestNumber(nums []int) int {
	ans, d := 0, 1000000
	for _, v := range nums {
		t := abs(v)
		if t < d || (t == d && v > ans) {
			ans, d = v, t
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

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