public class Solution {
public int HammingDistance(int x, int y) {
int distance = 0;
string sX = Convert.ToString(x, 2);
string sY = Convert.ToString(y, 2);
int maxLength = Math.Max(sX.Length, sY.Length);
//填充0,使两个字符串右对齐
sX = sX.PadLeft(maxLength, ‘0‘);
sY = sY.PadLeft(maxLength, ‘0‘);
for ( int i = 0 I MaxLength I ++) {
if (sX[i] != sY[i]) {
distance++;
}
}
return distance;
}
}
From for notes (Wiz)
end of the loop, only one number is not 0, the other number is 0, and at this point 0 corresponds to all the binary is 0, at this time as long as the number of the calculation is not 0 of binary 1 of the number.Improvement: The above uses three while, with one is enough:1 classSolution {2 Public:3 intHammingdistance (intXinty) {4 intdis =0;5 while(x | |y) {//or operation, as long as there is a not 0 for this loop6Dis + = (x 1) ^ (Y 1);7X >>=1;8Y >>=1;9 }Ten
The Hamming distance between and integers is the number of positions at which, the corresponding bits is different.Given integers x y and, calculate the Hamming distance.PS: To find the Hamming distance.Idea: The number of x and y binary xor 1Publicclasssolution{publicinthammingdistance (Intx, inty) {//stringx1= Integer.tobinarystring (x); //stringy1= Integer.tobinarystring (y);inttem=x^y; intcount=0;stringstr =integer.tobinarystring (TEM); for (Inti=0;iLeetcode
Problem Description:The Hamming distance between two integers refers to the number of different positions of these two numbers corresponding to bits.Give two integers x and y calculate the Hamming distance between them.Attention:0≤ x , y Example:input: x = 1, y = 4 output: 2 Explanation:1 (0 0 0 1 ) 4 (0 1 0 0) ↑ ↑ The above arrows pointed to the different positions of the corresponding bits.Method: 1 class Solution (object): 2 def Hammingdistance (self, x, y): 3 Span style
The Hamming distance between and integers is the number of positions at which, the corresponding bits is different.Given integers x y and, calculate the Hamming distance.Train of thought: Count two binary bitwise compare the number of different bits, first XOR, then the number of binary 1 of XOR. public class solution { public int hammingdistance (int x, int y) { int xor = x ^ y; int distance = 0; while (xor! = 0 ++; XOR = (xor-1); // " return distance; }}LeetCode-
fibs number of digits
problem:461
Time limit:1000ms
Memory limit:65536k
Description
The generalized Fibonacci sequence is defined as follows:
R0=a; R1= b; rn= uRn-1 + vRn-2 (n >= 3) Here a,b,u,v are real numbers, called RN for the generalized Fibonacci sequence. Now let you calculate the number of bits of the generalized Fibonacci sequence.
[Leetcode/javascript] 461.Hamming Distance
Topic
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
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. Analysis
The topic is very simple, given two 32-bit integers
Leetcode 461. Hamming Distance
Topics
The Hamming distance between and integers is the number of positions at which, the corresponding bits is different.
Given integers x and y, calculate the Hamming distance.
Solution:
Class Solution {public
:
int hammingdistance (int x, int y) {
int z = x^y;
int ans = 0;
while (z) {
ans++;
Z = z-1;
}
return ans;
}
;
Writing this blog is mainl
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Memo:
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Oracle learning-dmp file (table) Import and ExportPreface
We still use a lot of dmp files, which are used as the file format for oracle Import and Export tables. Today we will learn how to export and import dmp files.
Dmp File Export
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