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[Programming question] Replace spaces
[Programming question] Fibonacci sequence
[Programming question] Replace spaces
Please implement a function that replaces each space in a string with "%20". For example, when the string is We Are Happy., the replaced string is We%20Are%20Happy.
For interface type programming questions, you only need to follow the prompt code to implement this function;
The idea is as follows:
First, the number of spaces in the string is calculated, and the converted length of the string is determined by the number of spaces.
Every space becomes %20 is to encounter a space string length + 2
The string has two spaces, so add 4 to the string length. is end2;
If str[end1]!=' ' then str[end2]=str[end1]
If str[end1] == ' ' then str[end2--] = '0';str[end2--] = '2';str[end2--] = '%20';end1--;
code show as below:
class Solution {
public:
void replaceSpace(char* str, int length) {
//数空格
int space = 0;
char* p = str;
while (*p)
{
if (*p == ' ')
space++;
p++;
}
int newlen = length + space * 2;
int end1 = length - 1;
int end2 = newlen - 1;
while (end1 != end2)
{
if (str[end1] != ' ')
{
str[end2--] = str[end1--];
}
else
{
str[end2--] = '0';
str[end2--] = '2';
str[end2--] = '%';
end1--;
}
}
}
};[Programming question] Fibonacci sequence
The Fibonacci sequence is defined like this:
F[0] = 0
F[1] = 1
for each i ≥ 2: F[i] = F[i-1] + F[i-2]
Therefore, the Fibonacci sequence looks like : 0, 1, 1, 2, 3, 5, 8, 13, ..., the numbers in the Fibonacci sequence are called Fibonacci numbers. Give you an N, you want it to be a Fibonacci number, you can change the current number X to X-1 or X+1 at each step, now give you a number N to find the minimum number of steps required to become a Fibonacci number.
Give you a number N to find the minimum number of steps required to become a Fibonacci number
analyse as below:
Enter the number n, and first determine whether n belongs to the Fibonacci sequence through the loop. If it belongs to print 0, it means that 0 steps are required to make n a Fibonacci sequence; if n does not belong to the Fibonacci sequence, then use n and b is compared, if b>n, compare the distances from a to n and b to n, that is, compare the absolute values of (an) and (bn), if the absolute value of an is large, output the absolute value of bn, The opposite is true.
#define _CRT_SECURE_NO_WARNINGS
//斐波那契数列 求最小步数
#include<stdio.h>
#include<math.h>
int main()
{
int a = 0;
int b = 1;
int c = 1;
int n = 0;
scanf("%d", &n);
while (1)
{
if (b == n)
{
printf("0\n");
break;
}
else if(b > n)
{
if (abs(a - n) > abs(b - n))
{
printf("%d", abs(b - n));
}
else
{
printf("%d", abs(a - n));
}
break;
}
a = b;
b = c;
c = a + b;
}
return 0;
}