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# 70.Climbing-Stairs

## 题目描述

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

## 代码

### Approach #1 Brute Force

class Solution {
public int climbStairs(int n) {
return helper(0, n);
}
public int helper(int i, int n) {
if (i > n) return 0;
if (i == n) return 1;
return helper(i + 1, n) + helper(i + 2, n);
}
}

### Approach #2 Recursion with Memoization

class Solution {
public int climbStairs(int n) {
int memo[] = new int[n + 1];
return helper(0, n, memo);
}
private int helper(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] = helper(i + 1, n, memo) + helper(i + 2, n, memo);
return memo[i];
}
}

### Approach #3 Fibonacci Number

class Solution {
public int climbStairs(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;
}
}

### Approach #4 Dyanmic Programming

One can reach `i-th` step in one of the two ways:
1. 1.
Taking a single step from `i-1` step.
2. 2.
Taking a step of 2 from `i-2` step.
public class Solution {
public int climbStairs(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];
}
}

### Approach #5 Binets Method

Complexity Analysis
• Time complexity : O(logn). Traversing on logn bits.
• Space complexity : O(1). Constant space is used.
public class Solution {
public int climbStairs(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;
}
}

### Approach #5 Fibonacci Formula

Complexity Analysis
• Time complexity : O(logn). pow method takes nlogn time.
• Space complexity : O(1). Constant space is used.
public class Solution {
public int climbStairs(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);
}
}