Algorithm Exercises: Fibonacci Sequence
- Exercise 1 Implementation of matrix multiplication method of Fibonacci sequence
- Exercise 2: A person can go up one step at a time, or he can go up two steps. Return the number of ways for this person to go up N steps.
- Exercise 3 In the first year, there is a mature cow A on the farm. In each subsequent year: 1) Each mature cow will give birth to a cow. 2) Each new-born cow will give birth to a cow A. Three years to maturity 3) Each cow will never die Return the number of cows after N years
- Exercise 4: Given a number N, imagine all strings of length N composed of only two characters: 0 and 1. If in a string, there is a 1 immediately to the left of any 0 character, this string is considered to meet the standard. Returns how many strings meet the criteria
- Exercise 5: Use 1*2 tiles to fill the N*2 area and return the number of tile laying methods.
How to find matrix multiplication of Fibonacci sequence
1) The linear solution (O(N)) of the Fibonacci sequence is very easy to understand.
2) At the same time, using linear algebra, another representation can also be rewritten
| F(N), F(N-1) | = | F(2), F(1) | * N-2 power of a certain second-order matrix
3) Find this second-order matrix, and then find the N-2 power of this second-order matrix as quickly as possible
Recursive optimization similar to Fibonacci sequence
If a certain recursion, apart from the initial term, has the form
F(N) = C1 * F(N) + C2 * F(N-1) + … + Ck * F(Nk) (C1…Ck and k are constants)
And this recursive expression is strict and does not transfer with conditions.
Then there are optimizations similar to the Fibonacci sequence, and the time complexity can be optimized to O(logN)
Exercise 1 Implementation of matrix multiplication method of Fibonacci sequence
public static int f1(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return 1;
}
return f1(n - 1) + f1(n - 2);
}
public static int f2(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return 1;
}
int res = 1;
int pre = 1;
int tmp = 0;
for (int i = 3; i <= n; i++) {
tmp = res;
res = res + pre;
pre = tmp;
}
return res;
}
// O(logN)
public static int f3(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return 1;
}
// [ 1 ,1 ]
// [ 1, 0 ]
int[][] base = {
{ 1, 1 },
{ 1, 0 }
};
int[][] res = matrixPower(base, n - 2);
return res[0][0] + res[1][0];
}
public static int[][] matrixPower(int[][] m, int p) {
int[][] res = new int[m.length][m[0].length];
for (int i = 0; i < res.length; i++) {
res[i][i] = 1;
}
// res = 矩阵中的1
int[][] t = m;// 矩阵1次方
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
res = product(res, t);
}
t = product(t, t);
}
return res;
}
// 两个矩阵乘完之后的结果返回
public static int[][] product(int[][] a, int[][] b) {
int n = a.length;
int m = b[0].length;
int k = a[0].length; // a的列数同时也是b的行数
int[][] ans = new int[n][m];
for(int i = 0 ; i < n; i++) {
for(int j = 0 ; j < m;j++) {
for(int c = 0; c < k; c++) {
ans[i][j] += a[i][c] * b[c][j];
}
}
}
return ans;
}
public static void main(String[] args) {
int n = 19;
System.out.println(f1(n));
System.out.println(f2(n));
System.out.println(f3(n));
System.out.println("===");
}
Exercise 2: A person can go up one step at a time, or he can go up two steps. Return the number of ways for this person to go up N steps.
public static int s1(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return n;
}
return s1(n - 1) + s1(n - 2);
}
public static int s2(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return n;
}
int res = 2;
int pre = 1;
int tmp = 0;
for (int i = 3; i <= n; i++) {
tmp = res;
res = res + pre;
pre = tmp;
}
return res;
}
public static int s3(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return n;
}
int[][] base = { { 1, 1 }, { 1, 0 } };
int[][] res = matrixPower(base, n - 2);
return 2 * res[0][0] + res[1][0];
}
public static int[][] matrixPower(int[][] m, int p) {
int[][] res = new int[m.length][m[0].length];
for (int i = 0; i < res.length; i++) {
res[i][i] = 1;
}
// res = 矩阵中的1
int[][] t = m;// 矩阵1次方
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
res = product(res, t);
}
t = product(t, t);
}
return res;
}
public static void main(String[] args) {
int n = 19;
System.out.println(s1(n));
System.out.println(s2(n));
System.out.println(s3(n));
System.out.println("===");
}
Exercise 3 In the first year, there is a mature cow A on the farm. In each subsequent year: 1) Each mature cow will give birth to a cow. 2) Each new-born cow will give birth to a cow A. Three years to maturity 3) Each cow will never die Return the number of cows after N years
public static int c1(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2 || n == 3) {
return n;
}
return c1(n - 1) + c1(n - 3);
}
public static int c2(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2 || n == 3) {
return n;
}
int res = 3;
int pre = 2;
int prepre = 1;
int tmp1 = 0;
int tmp2 = 0;
for (int i = 4; i <= n; i++) {
tmp1 = res;
tmp2 = pre;
res = res + prepre;
pre = tmp1;
prepre = tmp2;
}
return res;
}
public static int c3(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2 || n == 3) {
return n;
}
int[][] base = {
{ 1, 1, 0 },
{ 0, 0, 1 },
{ 1, 0, 0 } };
int[][] res = matrixPower(base, n - 3);
return 3 * res[0][0] + 2 * res[1][0] + res[2][0];
}
public static int[][] matrixPower(int[][] m, int p) {
int[][] res = new int[m.length][m[0].length];
for (int i = 0; i < res.length; i++) {
res[i][i] = 1;
}
// res = 矩阵中的1
int[][] t = m;// 矩阵1次方
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
res = product(res, t);
}
t = product(t, t);
}
return res;
}
public static void main(String[] args) {
int n = 19;
System.out.println(c1(n));
System.out.println(c2(n));
System.out.println(c3(n));
System.out.println("===");
}
Exercise 4: Given a number N, imagine all strings of length N composed of only two characters: 0 and 1. If in a string, there is a 1 immediately to the left of any 0 character, this string is considered to meet the standard. Returns how many strings meet the criteria
public static int getNum1(int n) {
if (n < 1) {
return 0;
}
return process(1, n);
}
public static int process(int i, int n) {
if (i == n - 1) {
return 2;
}
if (i == n) {
return 1;
}
return process(i + 1, n) + process(i + 2, n);
}
public static int getNum2(int n) {
if (n < 1) {
return 0;
}
if (n == 1) {
return 1;
}
int pre = 1;
int cur = 1;
int tmp = 0;
for (int i = 2; i < n + 1; i++) {
tmp = cur;
cur += pre;
pre = tmp;
}
return cur;
}
public static int getNum3(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return n;
}
int[][] base = { { 1, 1 }, { 1, 0 } };
int[][] res = matrixPower(base, n - 2);
return 2 * res[0][0] + res[1][0];
}
public static int fi(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return 1;
}
int[][] base = { { 1, 1 },
{ 1, 0 } };
int[][] res = matrixPower(base, n - 2);
return res[0][0] + res[1][0];
}
public static int[][] matrixPower(int[][] m, int p) {
int[][] res = new int[m.length][m[0].length];
for (int i = 0; i < res.length; i++) {
res[i][i] = 1;
}
int[][] tmp = m;
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
res = product(res, tmp);
}
tmp = product(tmp, tmp);
}
return res;
}
// 两个矩阵乘完之后的结果返回
public static int[][] product(int[][] a, int[][] b) {
int n = a.length;
int m = b[0].length;
int k = a[0].length; // a的列数同时也是b的行数
int[][] ans = new int[n][m];
for(int i = 0 ; i < n; i++) {
for(int j = 0 ; j < m;j++) {
for(int c = 0; c < k; c++) {
ans[i][j] += a[i][c] * b[c][j];
}
}
}
return ans;
}
public static void main(String[] args) {
for (int i = 0; i != 20; i++) {
System.out.println(getNum1(i));
System.out.println(getNum2(i));
System.out.println(getNum3(i));
System.out.println("===================");
}
}