leetcode 70. Climbing Stairs
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题意
在每个台阶能走一步或两步,问到达第n个台阶能有多少种方式。
题解
和斐波那契序列一样,递归式为f(n) = f(n - 1) + f(n - 2),使用动态规划从底向上求解。
代码
class Solution {public: int climbStairs(int n) { int dp[n + 1]; dp[0] = dp[1] = 1; for(int i = 2; i <= n; i++) { dp[i] = dp[i - 1] + dp[i - 2]; } return dp[n]; }};
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