剑指 Offer 10- II. 青蛙跳台阶问题
记忆化递归
class Solution {private int[] memo;public int numWays(int n) {// f0=1;//0// f1=1;// f2=f1+f0;//1// f3=f2+f1;// f4=f2+f3;memo = new int[n + 1];Arrays.fill(memo, -1);return jump(n);}private int jump(int n) {if (memo[n] != -1) {return memo[n];}//用到缓存if (n == 1 || n == 0) {return 1;}memo[n] = (jump(n - 1) + jump(n - 2)) % 1000_000_007;return memo[n];}}
动态规划1
//动态规划public int numWays(int n) {if (n == 0 || n == 1) {return 1;}int[] dp = new int[n + 1];dp[1] = 1;dp[2] = 2;for (int i = 3; i <= n; i++) {dp[i] = (dp[i - 2] + dp[i - 1]) % 1000_000_007;}return dp[n];}
时间复杂度O(n),空间复杂度O(n)。
https://leetcode-cn.com/problems/qing-wa-tiao-tai-jie-wen-ti-lcof/solution/mian-shi-ti-10-ii-qing-wa-tiao-tai-jie-wen-ti-dong/
动态规划2
public int numWays(int n) {if (n == 0 || n == 1) {return 1;}int pre = 1, cur = 2;for (int i = 3; i <= n; i++) {int tmp = (pre + cur) % 1000_000_007;pre = cur;cur = tmp;}return cur;}
O(1)空间复杂度的动态规划。
