Multiple solutions of Fibonacci (Python or Java)
本章是用英文写的，作为或想成为一名优秀的攻城狮，习惯阅读英文文档将使你受益良多。例如更好的查看最新版的官方文档、与国外友人交流、等等 其实英文的生词也并不多,其中90％的英文都在代码里，当然这其中的精华也在代码里，代码相信大部分伙计还是都可以看懂.所以，请不要惊慌。对于English，让我们一起取克服它、习惯它、拥抱它。然后把它锤倒在地，相信你可以的。 GO， Go， GO 如果实在不行，各种页面翻译来一手。莫慌，这都是小场面，啥都不是事儿，好吧
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)
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);
}
}
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)
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);
}
}
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]
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]
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;
}
}
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]
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
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;
}
}
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)
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));
}
}
