Multiple solutions of Fibonacci (Python or Java)
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Violence law(Top-down)
It can be solved directly according to the known conditions (f (0) = 0, f (1) = 1 F(N) = F(N - 1) + F(N - 2), for N > 1)
Python Code
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class Solution:
def fib(self, N: int) -> int:
if N == 1 or N == 2: return N
return self.fib(N - 1) + self.fib(N - 2)
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Java Code
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class Solution {
public int fib(int N) {
if (N == 1 || N == 2) return 1;
return fib(N - 1) + fib(N - 2);
}
}
class Solution {
public int fib(int N) {
return N < 2 ? N : fib(N - 1) + fib(N - 2);
}
}
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Violence law add cache(Pruning)
We know that if we don’t do any processing, we will repeat too many calculations, which is very bad The processing idea will avoid repeated calculation
Python Code
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class Solution2:
@functools.lru_cache()
def fib(self, N: int) -> int:
if N <= 1: return N
else: return self.fib(N - 1) + self.fib(N - 2)
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Java Code
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class Solution {
private Integer[] cache = new Integer[31];
public int fib(int N) {
if (N <= 1) return N;
cache[0] = 0;
cache[1] = 1;
return memoize(N);
}
public int memoize(int N) {
if (cache[N] != null) return cache[N];
cache[N] = memoize(N-1) + memoize(N-2);
return memoize(N);
}
}
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Divide and conquer solution
Recursion, iteration, divide and conquer, backtracking, they do not have a clear distinction Recursion：The core idea is to govern separately and unify the officials
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class Solution:
def fib(self, N: int) -> int:
memo = {}
if N < 2: return N
if N-1 not in memo: memo[N-1] = self.fib(N-1)
if N-2 not in memo: memo[N-2] = self.fib(N-2)
return memo[N-1] + memo[N-2]
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Dynamic recursion(Bottom up)
Basic solutions
More initial value, continuous dynamic recursive
Python Code
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class Solution:
def fib(self, N: int) -> int:
if N < 2: return N
dp = [0 for _ in range(N + 1)]
dp[0], dp[1] = 0, 1
for i in range(2, N + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[- 1]
class Solution:
def fib(self, N: int) -> int:
if N == 0: return 0
memo = [0,1]
for _ in range(2,N+1):
memo = [memo[-1], memo[-1] + memo[-2]]
return memo[-1]
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Java Code
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class Solution {
public int fib(int N) {
if (N <= 1) return N;
if (N == 2) return 1;
int curr = 0, prev1 = 1, prev2 = 1;
for (int i = 3; i <= N; i++) {
curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
return curr;
}
}
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Use better base types (tuples) to improve performance
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class Solution:
def fib(self, N: int) -> int:
if N == 0: return 0
memo = (0,1)
for _ in range(2,N+1):
memo = (memo[-1], memo[-1] + memo[-2])
return memo[-1]
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Better solutions
Python Code
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class Solution:
def fib(self, N: int) -> int:
curr, prev1, prev2 = 0, 1, 1
for i in range(3, N + 1):
curr = prev1 + prev2
prev2 = prev1
prev1 = curr
return curr
class Solution5:
def fib(self, N: int) -> int:
prev, now = 0, 1
for i in range(N):
prev, now = now, now + prev
return prev
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Java Code
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class Solution {
public int fib(int N) {
if (N == 0) return 0;
if (N == 2 || N == 1) return 1;
int prev = 1, curr = 1;
for (int i = 3; i <= N; i++) {
int sum = prev + curr;
prev = curr;
curr = sum;
}
return curr;
}
}
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Mathematical conclusion method
Python Code
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class Solution:
def fib(self, N: int) -> int:
sqrt5 = 5 ** 0.5
fun = pow((1 + sqrt5) / 2, n + 1) - pow((1 - sqrt5) / 2, n + 1)
return int(fun / sqrt5)
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Java Code
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class Solution {
public int fib(int N) {
double sqrt5 = (1 + Math.sqrt(5)) / 2;
return (int)Math.round(Math.pow(sqrt5, N)/ Math.sqrt(5));
}
}
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文章来源: cuiqingcai.com，作者：Payne，版权归原作者所有，如需转载，请联系作者。
原文链接：cuiqingcai.com/10017.html
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