LeetCode 70 Climbing Stairs (crawling Stairs) (Dynamic Planning )(*)

LeetCode 70 Climbing Stairs (crawling Stairs) (Dynamic Planning )(*)
Translation

You are crawling a stair. It takes n steps to reach the top. You can crawl one or two steps each time. How many different methods do you have to climb to the top?

Original

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?

Analysis

To dynamically plan the basic questions, first set three variables for conversion:

int dp1 = 1, dp2 = 2, dpWay = 0;

According to the question, only one or two steps are allowed at a time. Therefore, when n is equal to 2, there are two steps: 1 + 1, 2.

if (n <= 1) return dp1;if (n == 2) return dp2;

Starting from 3, because you can directly obtain the result of its steps, you can directly write it:

while ((n--)-2) {}

The final change is as follows:

dpWay = dp1 + dp2;dp1 = dp2;dp2 = dpWay;

int climbStairsIter(int n,  int dpWay,int dp1, int dp2) {    if (n <= 1) return dp1;    if (n == 2) return dp2;    if ((n--) - 2) {        dpWay = dp1 + dp2;        dp1 = dp2;        dp2 = dpWay;        return climbStairsIter(n, dpWay, dp1, dp2);    }    else return dpWay;}int climbStairs(int n) {    return climbStairsIter(n, 0,1,2);}

Because the parameters here involve the execution of the preceding sequence, it is better to list them separately, but this seems a little less concise.

Code

class Solution {public:    int climbStairs(int n) {        int dp1 = 1, dp2 = 2, dpWay = 0;        if (n <= 1) return dp1;        if (n == 2) return dp2;        while ((n--) - 2) {            dpWay = dp1 + dp2;            dp1 = dp2;            dp2 = dpWay;        }        return dpWay;    }};