lc.509 斐波那契
- DP 滚动数组
class Solution {
public:int fib(int n) {
int f[31];f[0]=0;f[1]=1;for(int i=2;i<=n;i++){
f[i]=f[i-1]+f[i-2];}return f[n];}
};
- 矩阵快速幂 快速幂算法O(logn)
class Solution {
public:int fib(int n) {
if (n < 2) {
return n;}vector<vector<int>> q{
{
1, 1}, {
1, 0}};vector<vector<int>> res = matrix_pow(q, n - 1);return res[0][0];} //返回res[0][0]==Fn//由于f0=0,f1=1,res*[1;0]等于res，所以没必要再乘[1;0]vector<vector<int>> matrix_pow(vector<vector<int>>& a, int n) {
vector<vector<int>> ret{
{
1, 0}, {
0, 1}};while (n > 0) {
if (n & 1) {
ret = matrix_multiply(ret, a); //矩阵与矩阵[1,0;0,1]相乘等于其本身}n >>= 1;a = matrix_multiply(a, a);}return ret;} //快速幂实现a^n 函数matrix_powvector<vector<int>> matrix_multiply(vector<vector<int>>& a, vector<vector<int>>& b) {
vector<vector<int>> c{
{
0, 0}, {
0, 0}};for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];}}return c;} //实现两个矩阵a,b相乘 函数matrix_multiply
};我只觉得妙不堪言
- 数学方法，通项公式
class Solution {
public:int fib(int n) {
double sqrt5 = sqrt(5);double fibN = pow((1 + sqrt5) / 2, n) - pow((1 - sqrt5) / 2, n);return round(fibN / sqrt5);}
};lc.1137 泰波那契
Tn=Tn-1+Tn-2+Tn-3
其中，T1=0，T2=1，T3=1
- DP
class Solution {
public:int tribonacci(int n) {
if (n == 0){
return 0;}if (n <= 2) {
return 1;}int p = 0, q = 0, r = 1, s = 1;for (int i = 3; i <= n; ++i) {
p = q;q = r;r = s;s = p + q + r;}return s;}
};
- 快速幂
class Solution {
public:int tribonacci(int n) {
if (n == 0) {
return 0;}if (n <= 2) {
return 1;}vector<vector<long>> q = {
{
1, 1, 1}, {
1, 0, 0}, {
0, 1, 0}};vector<vector<long>> res = pow(q, n);return res[0][2];}vector<vector<long>> pow(vector<vector<long>>& a, long n) {
vector<vector<long>> ret = {
{
1, 0, 0}, {
0, 1, 0}, {
0, 0, 1}};while (n > 0) {
if ((n & 1) == 1) {
ret = multiply(ret, a);}n >>= 1;a = multiply(a, a);}return ret;}vector<vector<long>> multiply(vector<vector<long>>& a, vector<vector<long>>& b) {
vector<vector<long>> c(3, vector<long>(3));for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j];}}return c;}
};和 lc.509 差不多，依然是构造了multiply和pow两个函数来使用快速幂，二维变三维。
注意到可以用函数的构造，也同样可以使用字符重载。