A little bit confused about the problem description

  • 0
    int climbStairs(int n){
    int dp[n + 1];
    fill_n(dp, n + 1, 0);
    dp[0] = 0;
    dp[1] = 1;
    dp[2] = 2;
    for(int i = 3; i <= n; ++i)
        dp[i] = dp[i - 1] + dp[i - 2];
    return dp[n];}

    Above is one of the desired solutions. The problem is simple if you consider "climb to the top" is an "exact reach" to top. However, at first glance, I think it's OK to climb "beyond" the top. I mean, for example, when n = 2, there are 3 ways of climbing: 1-1, 2 and 1-2. If the description wants to use "climbing stair" to visualize, my understanding might be more close to the reality.

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