💯 Daily LeetCode Challenge Day_05 - Hamming Distance

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문제

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4
Output: 2
Explanation: 1 (0 0 0 1) 4 (0 1 0 0) ↑ ↑ The above arrows point to positions where the corresponding bits are different.

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Swift 는 String(x, radix: 2) 로 진수 변환을 할 수 있다.

2진수로 바꿔준 뒤 pad 함수를 이용하여 String 길이가 긴 이진수에 맞춰 0문자를 넣어줘 문자열 길이를 같게 만든다.

그리고 문자가 다르면 answer += 1 하며 정답을 찾는다.

class Solution {
    func hammingDistance(_ x: Int, _ y: Int) -> Int {
        //decimal to binary
        var binX = String(x, radix: 2)
        var binY = String(y, radix: 2)
        //make same digit
        if binX.count > binY.count{
            binY = pad(binY, binX.count)
        }else{
            binX = pad(binX,binY.count)
        }
        //compare
        var answer = 0
        for i in 0..<binX.count{
            let index = binX.index(binX.startIndex, offsetBy: i)
            if binX[index] != binY[index] {
                answer += 1
            }
        }
        
        return answer
    }
    func pad(_ string : String,_ toSize: Int) -> String {
        var padded = string
        for _ in 0..<(toSize - string.count) {
            padded = "0" + padded
        }
        return padded
    }
}