70. Climbing Stairs
Easy
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
package leetcode.easy;public class ClimbingStairs {
@org.junit.Test
public void test() {
int n1 = 2;
int n2 = 3;
System.out.println(climbStairs1(n1));
System.out.println(climbStairs1(n2));
System.out.println(climbStairs2(n1));
System.out.println(climbStairs2(n2));
System.out.println(climbStairs3(n1));
System.out.println(climbStairs3(n2));
System.out.println(climbStairs4(n1));
System.out.println(climbStairs4(n2));
System.out.println(climbStairs5(n1));
System.out.println(climbStairs5(n2));
System.out.println(climbStairs6(n1));
System.out.println(climbStairs6(n2));
}public int climbStairs1(int n) {
return climb_Stairs(0, n);
}public int climb_Stairs(int i, int n) {
if (i > n) {
return 0;
}
if (i == n) {
return 1;
}
return climb_Stairs(i + 1, n) + climb_Stairs(i + 2, n);
}public int climbStairs2(int n) {
int[] memo = new int[n + 1];
return climb_Stairs(0, n, memo);
}public int climb_Stairs(int i, int n, int memo[]) {
if (i > n) {
return 0;
}
if (i == n) {
return 1;
}
if (memo[i] > 0) {
return memo[i];
}
memo[i] = climb_Stairs(i + 1, n, memo) + climb_Stairs(i + 2, n, memo);
return memo[i];
}public int climbStairs3(int n) {
if (n == 1) {
return 1;
}
int[] dp = new int[n + 1];
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}public int climbStairs4(int n) {
if (n == 1) {
return 1;
}
int first = 1;
int second = 2;
for (int i = 3; i <= n; i++) {
int third = first + second;
first = second;
second = third;
}
return second;
}public int climbStairs5(int n) {
int[][] q = { { 1, 1 }, { 1, 0 } };
int[][] res = pow(q, n);
return res[0][0];
}public int[][] pow(int[][] a, int n) {
int[][] ret = { { 1, 0 }, { 0, 1 } };
while (n > 0) {
if ((n & 1) == 1) {
ret = multiply(ret, a);
}
n >>= 1;
a = multiply(a, a);
}
return ret;
}public int[][] multiply(int[][] a, int[][] b) {
int[][] c = new int[2][2];
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;
}public int climbStairs6(int n) {
double sqrt5 = Math.sqrt(5);
double fibn = Math.pow((1 + sqrt5) / 2, n + 1) - Math.pow((1 - sqrt5) / 2, n + 1);
return (int) (fibn / sqrt5);
}
}
