509. 斐波那契数(动态规划)
斐波那契数 (通常用 F(n) 表示)形成的序列称为 斐波那契数列 。该数列由 0 和 1 开始,后面的每一项数字都是前面两项数字的和。也就是:
F(0) = 0,F(1) = 1
F(n) = F(n - 1) + F(n - 2),其中 n > 1
给定 n ,请计算 F(n) 。
示例 1:
输入:n = 2
输出:1
解释:F(2) = F(1) + F(0) = 1 + 0 = 1
示例 2:
输入:n = 3
输出:2
解释:F(3) = F(2) + F(1) = 1 + 1 = 2
示例 3:
输入:n = 4
输出:3
解释:F(4) = F(3) + F(2) = 2 + 1 = 3
提示:
0 <= n <= 30
题解一(动规五部曲)
class Solution {
public int fib(int n) {
if (n <= 1) {
return n;
}
int[] dp = new int[n + 1];
dp[0] = 0;
dp[1] = 1;
for (int i = 2; i < dp.length; ++i) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
}
解析
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题解二(动规空间优化)
class Solution {
public int fib(int n) {
int a = 0, b = 1, sum = 0;
if (n <= 1) {
return n;
}
for (int i = 2; i < n + 1; ++i) {
sum = a + b;
a = b;
b = sum;
}
return sum;
}
}
解析
题解三(递归法)
class Solution {
public int fib(int n) {
if (n <= 1) {
return n;
}
return fib(n - 1) + fib(n - 2);
}
}