And we can just finish this problem with almost the same way haha!
C# Solution
Solution1 – using three variables
public class Solution {
public int Fib(int n) {
int first = 0;
int second = 1;
int tmp = 0;
if(n<2)
{
return n;
}
else
{
for(int i =2; i<=n;i++)
{
tmp = second;
second+=first;
first = tmp;
}
}
return second;
}
}
Runtime : 25ms、24ms、28ms (faster than 40~70%)
Solution2 – using one array
public class Solution {
public int Fib(int n) {
int[] arr = new int[31];
arr[0] = 0;
arr[1] = 1;
for(int i = 2; i<=n; i++)
{
arr[i] = arr[i-1]+arr[i-2];
}
return arr[n];
}
}
We can the array size as 31 is because (Constraints : 0<=n<=30)
The maximum of n is 30, and because index start from zero, so it needs to plus one.
Runtime : 23ms、24ms、23ms (faster than 70~80%)
⭐ If the range of n is larger, it is recommended to dynamically set the array length.
➡ This can help avoid excessive memory usage.
Java Solution
Solution1 – using one array
class Solution {
public int fib(int n) {
int[]arr = new int[31];
arr[0] = 0;
arr[1] = 1;
for(int i = 2; i<=n; i++)
{
arr[i] = arr[i-1]+arr[i-2];
}
return arr[n];
}
}
Runtime : 0ms、0ms、0ms
Python3 Solution
Solution1 – using list
class Solution:
def fib(self, n: int) -> int:
arr = [0,1]
for i in range(2,n+1):
arr.append(arr[i-1]+arr[i-2])
return arr[n]
JavaScript Solution
Solution1
/**
* @param {number} n
* @return {number}
*/
var fib = function(n) {
arr = [0,1];
for(let i = 2; i<=n; i++)
{
arr[i] = arr[i-1]+arr[i-2];
//arr.push(arr[i-1]+arr[i-2]) //you can choose which one to use
}
return arr[n];
};
Conclusion
🧡If my solution helps, that is my honor!
🧡You can support me by clicking some ad or sharing my posts, Thanks a lot
✅If you got any problem about the explanation, please feel free to let me know
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