70. 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
//这道题其实就是斐波那契数列,用递归会超时 //Approach: Recursive Time: O(n!), Space: O(1) Tim Limited Exceed public int climbStairs(int n) { if (n <= 2) { return n; } return climbStairs(n - 1) + climbStairs(n - 2); } //Approach: Iteration Time: O(n) //1. Your runtime beats 2.35 % of java submissions. public int climbStairs(int n) { if (n <= 2) { return n; } int first = 1; int second = 2; int result = 0; for (int i = 3; i <= n; i++) { result = first + second; first = second; second = result; } return result; } //2. Your runtime beats 48.13 % of java submissions. public int climbStairs(int n) { if (n <= 1) return 1; int[] dp = new int[n]; dp[0] = 1; dp[1] = 2; for (int i = 2; i < n; ++i) { dp[i] = dp[i - 1] + dp[i - 2]; } return dp[n - 1]; } //3. Your runtime beats 100.00 % of java submissions public int climbStairs(int n) { int a = 1, b = 1; while (n > 0) { b += a; a = b - a; n--; } return a; }