70. Climbing Stairs (Easy)
https://leetcode.com/problems/climbing-stairs/
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
Solutions
1.
class Solution {
public int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}
if (n == 2) {
return 2;
}
// there are n stairs, actually, the walking path start from 0, the n
int[] pathWay = new int[n];
// stand still, 1 possibility
pathWay[n - 1] = 1;
// from n-2 to n, 2 possibility
pathWay[n - 2] = 2;
for (int i = n - 3; i >= 0; i--) {
pathWay[i] = pathWay[i + 1] + pathWay[i + 2];
}
return pathWay[0];
}
}
2.
class Solution {
// pay attention that there is n steps, suppose the topmost stair numbered 0,
// which means your starting stair's number is n, rather than n-1
int[] pathWay;
public int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}
if (n == 2) {
return 2;
}
pathWay = new int[n + 1];
Arrays.fill(pathWay, -1);
pathWay[1] = 1;
pathWay[2] = 2;
// set off from n to 0
return walkThrough(n);
}
public int walkThrough(int n) {
if (n <= 0) {
return 0;
}
if (pathWay[n] >= 0) {
return pathWay[n];
}
int p1 = walkThrough(n - 1);
int p2 = walkThrough(n - 2);
pathWay[n] = p1 + p2;
return p1 + p2;
}
}