Climbing Stairs (leetcode)

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?

使用动态规划的方法解决此问题：

位置n 只能由 位置(n- 1)走一步，或者位置 (n - 2)走两步到达，即：

step [n] = step[n-1] + step[n - 2];

1> 方法一：

   时间复杂度 O(n);

  空间复杂度 0(n);

class Solution {public:    int climbStairs(int n) {        //step[n] = step[n-1] + step[n -2]        if (n <= 2) {            return n;        }        int *step = new int[n + 1];        step[0] = 0;        step[1] = 1;        step[2] = 2;            for (int i = 3; i <= n; ++i) {            step[i] = step[i -1] + step[i - 2];        }        unsigned int retval = step[n];        delete [] step;        return retval;    } };

2> 方法2

   时间复杂度  : O(n)

   空间复杂度   : O(1)

class Solution {public:    int climbStairs(int n) {        //step[n] = step[n-1] + step[n -2]    if (n <= 2) {    return n;    }    int step_curr = 0;    //step[n]    int step_pre = 2;     //step[n - 1]    int step_pre_pre = 1; //step[n - 2]    for (int i = 0; i < n - 2; ++i) {        step_curr = step_pre + step_pre_pre;        step_pre_pre = step_pre;        step_pre = step_curr;    }    return step_curr;      }};