70. Climbing Stairs dynamic programming

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.

首先是递归的算法，但是会超时

int climbStairs(int n) {    if (n <= 0)return 0;    else if (n == 1)return 1;    else if (n == 2)return 2;    return climbStairs(n - 1) + climbStairs(n - 2);}

接着用一个数组存储到第i步有多少种走法，但是这样的空间复杂度为O(n)；

int climbStairs(int n) {    vector<int>my;    if (n <= 0)return 0;    else if (n == 1)return 1;    else if (n == 2)return 2;    my.push_back(0);    my.push_back(1);    my.push_back(2);    for (int i = 3; i <= n; i++){        my.push_back(my[i - 1] + my[i - 2]);    }    return my[n];}

最后，将空间复杂度降为O（1）；

int climbStairs(int n) {    if (n == 0)return 0;    if (n == 1)return 1;    if (n == 2)return 2;    int pre = 1, cur = 2, next = 0;    for (int i = 3; i <= n; i++){        next = pre + cur;        pre = cur;        cur = next;    }    return cur;}