LeetCode 70. Climbing Stairs

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本文探讨了经典的爬楼梯问题，提供了两种解决方案：一种使用动态规划的经典方法，另一种则是更为简洁高效的迭代方法。通过这两种方法，展示了如何计算出达到楼梯顶部的不同方式的数量。

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?

Classical DP problem

int climbStairs(int n) {
        vector<int> dp(n + 1, 0);
        dp[1] = 1;
        dp[0] = 1;
        for(int i = 2; i <= n; ++i) {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }

A much better version.

int climbStairs(int n) {
    int cn1 = 1;
    int cn2 = 2;
    if(n <= 0) return 0; if(n == 1) return 1;
    for(int i = 3; i <= n; ++i) {
       cn1 += cn2;
       swap(cn1, cn2);
    }
    return cn2;
}