LeetCode #461 HammingDistance. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
바로 전에 bitset 을 새로 알게되었어서,,, 그래서 쉬웠다.. 100% 퍼포먼스
I learned bitset just right before this problem,,, so it was hella easy.. 100% performance
2019/02/24 - [Programming/C++] - How to use bitset, bitwise operator in c++, 비트셋, 비트연산자 사용하는 방법, 이진수 변환 방법, How to convert decimal to binary, binary to decimal
LeetCode #461
Q.
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 < 2^31
해밍거리는 두 정수 사이의 상응하는 자리 bit 가 다른 것의 개수를 나타낸다.
두개의 정수 x, y 가 있을 때, 해밍거리를 계산해라.
노트 :
0 <= x, y < 2^31
e.g.
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.
상응하는 두개의 자리값이 다르다.
Process
// Process
//1. Input int x, y
//2. Make xBit and yBit
//3. Iterate from begin to the end bit
// 3.1. Compare if it's diff or same
// 3.1.1. If diff, count
//4. Return count
// 처리과정
//1. 정수 x, y 를 입력받는다.
//2. xBit 과 yBit bitset 을 만든다.
//3. 시작부터 끝까지 반복한다.
// 3.1. 해당 bit 가 다른지 확인해서
// 3.1.1. 다르면 개수 센다.
//4. 개수 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
int hammingDistance(int x, int y) {
bitset<32> xBit(x);
bitset<32> yBit(y);
int count = 0;
for (int i = 0; i < xBit.size(); ++i)
if (xBit[i] == yBit[i])
++count;
return 32-count;
}
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