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#
Solution 1#
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| class Solution:
def findClosestNumber(self, nums: List[int]) -> int:
ans, d = 0, inf
for x in nums:
if (y := abs(x)) < d or (y == d and x > ans):
ans, d = x, y
return ans
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| class Solution {
public int findClosestNumber(int[] nums) {
int ans = 0, d = 1 << 30;
for (int x : nums) {
int y = Math.abs(x);
if (y < d || (y == d && x > ans)) {
ans = x;
d = y;
}
}
return ans;
}
}
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| class Solution {
public:
int findClosestNumber(vector<int>& nums) {
int ans = 0, d = 1 << 30;
for (int x : nums) {
int y = abs(x);
if (y < d || (y == d && x > ans)) {
ans = x;
d = y;
}
}
return ans;
}
};
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| func findClosestNumber(nums []int) int {
ans, d := 0, 1<<30
for _, x := range nums {
if y := abs(x); y < d || (y == d && x > ans) {
ans, d = x, y
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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| function findClosestNumber(nums: number[]): number {
let [ans, d] = [0, 1 << 30];
for (const x of nums) {
const y = Math.abs(x);
if (y < d || (y == d && x > ans)) {
[ans, d] = [x, y];
}
}
return ans;
}
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