Dynamic Programming Solution Using Java

  • 9
     * Ideas:
     * Use Dynamic Programming,
     * for each step, the stair could ether combine with the previous one or as a single step.
     * Ways to climb to ith stair is W(i) = W(i-1) + W(i-2)
     * where W(i-1) is when the ith stair is as a single step
     * and W(i-2) is when the ith stair is paired with the previous one.
    public int climbStairs(int n) {
            int[] tmp = new int[n];
            if (n < 2){
                return 1;
            tmp[0] = 1;
            tmp[1] = 2;
            for (int i = 2; i < n; i++){
                tmp[i] = tmp[i-1] + tmp[i-2];
            return tmp[n-1];

  • 0

    why my java recursive code has "Status: Time Limit Exceeded" error? is it same as OP's array code?

    public class Solution {
            public int climbStairs(int n) {
                 if (n==1) return 1;
               else if (n==2) return 2;
            else return  climbStairs(n-1)+climbStairs(n-2);

  • 0

    Recursive method usually takes much longer time than the loop.
    It has to call itself multiple times. There will be a very large call and return stack when the input number is big.

  • 0

    The regular recursion method will call climbStairs(n) for the same n several times. You can store the results for each n in a collection, and reuse them afterwards. That will be an dynamic programming recursion.

  • 0

    @houzhaowei I think it should be return n for the corner case when n is less or equal to 2. For n = 2, it could be two of one step OR one of two steps.

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