斐波那契数列、跳台阶、变态跳台阶、矩形覆盖

#include<iostream>#include<math.h>using namespace std;//斐波那契数列class Solution{public:int Fibonacci1(int n){/**第一种解法：递归————从上往下计算，重复计算太多，效率太低，不能满足时间要求*/if (n == 0)return 0;if (n == 1)return 1;return Fibonacci1(n - 1) + Fibonacci1(n - 2);}int Fibonacci2(int n){/**第二种解法：从下往上计算，减少重复计算*/int n0 = 0, n1 = 1;int ret;if (n == 0)ret = 0;if (n == 1)ret = 1;for (int i = 2; i <= n; i++){ret = n0 + n1;n0 = n1;n1 = ret;}return ret;}};//扩展1：青蛙跳台阶————一次可以跳1级，也可以跳2级，求跳N级的跳法种数。class Solution2{public:int jumpFloor1(int n){/**n=1：只有1种跳法；*n=2：两种跳法；*n>2：假设跳N级的跳法有f(n)种，*         （1）第一次跳，跳1级，则剩下N-1级的跳法有f(n-1)种；*         （2）第一次跳，跳2级，则剩下N-2级的跳法有f(n-2)种；*     所以：f(n) = f(n-1) + f(n-2)。*///if (n == 0 || n == 1)if (n == 1)return 1;if (n == 2)return 2;return jumpFloor1(n - 1) + jumpFloor1(n - 2);}int jumpFloor2(int number) {int n0 = 1, n1 = 2;int ret;if (number == 1)ret = 1;if (number == 2)ret = 2;for (int i = 3; i <= number; i++){ret = n0 + n1;n0 = n1;n1 = ret;}return ret;}int jumpFloor3(int number) {/**第3种解法：滚动数组*/int F[3] = { 0, 1, 2 };if (number <= 2){return F[number];}for (int i = 3; i <= number; i++){F[0] = F[1];F[1] = F[2];F[2] = F[0] + F[1];}return F[2];}};//扩展2：青蛙变态跳台阶————一次可以跳1级，也可以跳2级.....，也可以跳上N级台阶，求跳N级的跳法种数。class Solution3{public:int jumpFloorII(int number) {/**第1种解法：递归————从顶往下计算*     f(n) = f(0)+f(1)+f(2)+f(3)+......+f(n-2)+f(n-1) ———— 第一次跳n，第一次跳n-1......*又   f(n-1) = f(0)+f(1)+f(2)+f(3)+......+f(n-2)*所以 f(n) - f(n-1) = f(n-1) -----> f(n) = 2 * f(n-1)*/if (number == 1)return 1;return 2 * jumpFloorII(number - 1);}int jumpFloorII1(int number) {/**第2种解法：从下往顶计算*最后剩下0个台阶，暂且定为0，直接跳n个台阶上来，显然只有一种方法（也就是说，对于任何一个“N”，我们都可以直接跳“N”，所以我们每次循环首先自加1就行了）*最后剩下1个台阶，那么共有（第n-1个台阶的方法数）种；*最后剩下2个台阶，共有(第n-2个台阶的方法数)种；*....*最后剩下n-1个台阶，只有一种方法。*累加上面每一种的方法，就是跳到第n阶台阶的总的跳法种数*/long long int arr[100] = { 0, 1 };      //限定最多跳100级for (int i = 2; i < 100; i++){int j = i - 1;arr[i]++;      //直接跳跃到本身的"N"while (j){arr[i] = arr[i] + arr[j];j--;}}return arr[number];}int jumpFloorII2(int number) {/**第3种解法：用数学归纳法证明f(n)=2^(n-1)。*/return pow(2, number - 1);}int jumpFloorII3(int number) {/**第4种解法：滚动数组*/int F[2] = { 0, 1 };if (number < 2)return F[number];for (int i = 2; i <= number; i++){F[0] = F[1];F[1] = 2 * F[0];}return F[1];}};//扩展3：矩形覆盖————用number个2*1的矩形去覆盖更大的矩形（number*2），求总共的方法数。class Solution4 {public:int rectCover(int number) {if (number <= 0 || number == 1)return 1;if (number == 2)return 2;return rectCover(number - 1) + rectCover(number - 2);}int rectCover1(int number) {if (number == 1)return 1;if (number == 2)return 2;int n1 = 1, n2 = 2, tmp = 0;for (int i = 3; i <= number; i++){tmp = n1 + n2;n1 = n2;n2 = tmp;}return tmp;}};void main(){Solution4 s;cout << s.rectCover1(5) << endl;}