标签：内存 改进 -- 解决问题 end pre 结果 数据 use
《程序员代码面试指南--IT名企算法与数据结构题目最优解》 左程云 著
斐波拉契数列问题的递归和动态规划
【题目】：
给定整数N,返回斐波拉契数列的第N项。
补充问题1：给定整数N，代表台阶数，一次可以跨2个或者1个台阶，返回有多少种走法。
补充问题2：假设农场中成熟的母牛每年只会生产1头小母牛，并且永远不会死。第一年农场只有1只成熟的母牛，
从第2年开始，母牛开始生产小母牛。每只小母牛3年后成熟又可以生产小母牛。给定整数N，求出N年后牛的数量。
【举例】
斐波拉契数列
f(0)=0, f(1)=1,f(2)=f(1)+f(0),f(3)=f(2)+f(1),
f(N)=f(N-1)+f(N-2) ///后一项是前两项之和
【举例】
青蛙跳台阶：
N=1，1种跳法
N=2，2种跳法
如果台阶N级，最后跳上第N级的情况，要么是从N-2台阶直接跨2级台阶，要么是从N-1级台阶跨1级台阶。
所以台阶有N级的方法为 f(N-2) + f(N-1)
【举例】
母牛问题：
第1年，a
第2年，a,b //b出生了
第3年，a,b,c
第4年，a,b,c,d
第5年，a,b,c,d, e,[b‘] //b成熟了
第6年，a,b,c,d,e,[b‘] f, b‘‘, c‘ //c,b成熟了
第N年，S(N) = S(N-1)+S(N-3)
问题进阶：可以用更快的时间复杂度解决问题。
最快可达到O(log(N))，需要用到矩阵乘法、状态矩阵、加速矩阵乘法等等数学知识。
限于本人数学功底薄弱，无法达到这个境界，看不懂。这里只是提出来。
#include <iostream> #include <stack> #include <exception> using namespace std;
////时间复杂度是O(2^N)，指数级的，不推荐 int fibnac(int n) { if (n < 0) { throw new std::exception("Invalid para"); return -1; } if (n == 0) { return 0; } if (n <= 2) { return 1; } return fibnac(n - 1) + fibnac(n - 2); }
//改进后的算法。时间复杂度为O(N)。减少重复的计算。 //自下而上分析，自上而下解决。递归的思路分析，循环的思路写代码。 unsigned long long fibnac2(int n) { if (n < 0) { throw new std::exception("Invalid para"); } unsigned long long *fibNums = new unsigned long long[n + 1]; fibNums[0] = 0; fibNums[1] = 1; fibNums[2] = 1; fibNums[3] = 2; for (unsigned long long i = 2; i <= n; i++) { fibNums[i] = fibNums[i - 1] + fibNums[i - 2]; } return fibNums[n]; }
////时间复杂度O(N) ////如果想节约内存空间，可以只声明3个变量来存储计算结果，进行累加计算 unsigned long long fibnac3(int n) { if (n < 0) { throw new std::exception("Invalid para"); } unsigned long long pre = 0; unsigned long long now = 1; if (n == 0) { return 0; } else if (n == 1 || n == 2) { return 1; } unsigned long long preTemp = 0; for (int i = 2; i <= n; i++) { preTemp = now; now += pre ; pre = preTemp; } return now; }
////时间复杂度O(N) ////青蛙跳台阶，利用临时变量累加 unsigned long long jump(int n) { if (n < 0) { throw new std::exception("Invalid para"); } if (n == 0) { return 0; } if (n <= 2) { return n; } unsigned long long llPre1 = 1;//n==1 unsigned long long llNow = 2;//n==2 unsigned long long llTemp = 0; for (int i = 3; i <= n; i++)//n>=3 { llTemp = llNow; llNow += llPre1;//n>=3 llPre1 = llTemp; } return llNow; }
////时间复杂度O(N) ////青蛙跳台阶，利用数组实现累加（理解更方便，但是需要一定的额外内存空间） unsigned long long jump2(int n) { if (n < 0) { throw new std::exception("Invalid para"); } if (n == 0) { return 0; } if (n <= 2) { return n; } unsigned long long *jumpArr = new unsigned long long[n + 1]; jumpArr[0] = 0; jumpArr[1] = 1; jumpArr[2] = 2; for (int i = 3; i <= n; i++)//n>=3 { jumpArr[i] = jumpArr[i-1] + jumpArr[i-2]; } return jumpArr[n]; }
////时间复杂度O(N) ///母牛产子问题 unsigned long long cowBron(int n) { if (n < 0) { throw new std::exception("Invalid para"); } if (n == 0) { return 0; } else if (n == 1 || n == 2 || n==3) { return n; } unsigned long long llprepre = 1; unsigned long long llpre = 2; unsigned long long llRes = 3; unsigned long long lltemp1 = 0; unsigned long long lltemp2 = 0; for (int i = 4; i <= n; i++) { lltemp1 = llpre; lltemp2 = llRes; llRes = llRes + llprepre; llprepre = lltemp1; llpre = lltemp2; } return llRes; }
////时间复杂度O(N) ///母牛产子问题 unsigned long long cowBron2(int n) { if (n < 0) { throw new std::exception("Invalid para"); } if (n == 0) { return 0; } else if (n == 1 || n == 2 || n == 3) { return n; } unsigned long long *llcowCounts = new unsigned long long[n + 1]; llcowCounts[0] = 0; llcowCounts[1] = 1; llcowCounts[2] = 2; llcowCounts[3] = 3; for (int i = 4; i <= n; i++) { llcowCounts[i] = llcowCounts[i - 1] + llcowCounts[i - 3]; } return llcowCounts[n]; }
//==========================测试用例==================================== void test1() { cout << fibnac(0) << endl;; cout << fibnac(1) << endl;; cout << fibnac(2) << endl;; cout << fibnac(10) << endl;; cout << fibnac(100) << endl;; //cout << fibnac(-5) << endl;; } void test2() { cout << fibnac2(0) << endl; cout << fibnac2(1) << endl; cout << fibnac2(2) << endl; cout << fibnac2(10) << endl; cout << fibnac2(11) << endl; cout << fibnac2(12) << endl; cout << fibnac2(50) << endl;///数字太大越界了 // cout << fibnac2(-5) << endl; } void test3() { cout << "----------------test3-------------------" << endl; cout << 0<<": "<<jump(0) << endl; cout << 1 << ": " << jump(1) << endl; cout << 2 << ": " << jump(2) << endl; cout << 3 << ": " << jump(3) << endl; cout << 4 << ": " << jump(4) << endl; cout << 5 << ": " << jump(5) << endl; cout << 14 << ": " << jump(14) << endl; cout << 15 << ": " << jump(15) << endl; cout << 16 << ": " << jump(16) << endl; cout << "-----------jump2-----------" << endl; cout << 0 << ": " << jump2(0) << endl; cout << 1 << ": " << jump2(1) << endl; cout << 2 << ": " << jump2(2) << endl; cout << 3 << ": " << jump2(3) << endl; cout << 4 << ": " << jump2(4) << endl; cout << 5 << ": " << jump2(5) << endl; cout << 14 << ": " << jump2(14) << endl; cout << 15 << ": " << jump2(15) << endl; cout << 16 << ": " << jump2(16) << endl; } void test4() { cout << "----------------test4-------------------" << endl; cout << 0 << ": " << cowBron(0) << endl; cout << 1 << ": " << cowBron(1) << endl; cout << 2 << ": " << cowBron(2) << endl; cout << 3 << ": " << cowBron(3) << endl; cout << 4 << ": " << cowBron(4) << endl; cout << 5 << ": " << cowBron(5) << endl; cout << 6 << ": " << cowBron(6) << endl; cout << 7 << ": " << cowBron(7) << endl; cout << 8 << ": " << cowBron(8) << endl; cout << 9 << ": " << cowBron(9) << endl; cout << 14 << ": " << cowBron(14) << endl; cout << 15 << ": " << cowBron(15) << endl; cout << 16 << ": " << cowBron(16) << endl; cout << "-----------cowBron2-----------" << endl; cout << 0 << ": " << cowBron2(0) << endl; cout << 1 << ": " << cowBron2(1) << endl; cout << 2 << ": " << cowBron2(2) << endl; cout << 3 << ": " << cowBron2(3) << endl; cout << 4 << ": " << cowBron2(4) << endl; cout << 5 << ": " << cowBron2(5) << endl; cout << 6 << ": " << cowBron2(6) << endl; cout << 7 << ": " << cowBron2(7) << endl; cout << 8 << ": " << cowBron2(8) << endl; cout << 9 << ": " << cowBron2(9) << endl; cout << 14 << ": " << cowBron2(14) << endl; cout << 15 << ": " << cowBron2(15) << endl; cout << 16 << ": " << cowBron2(16) << endl; } int main() { //test1(); test2(); test3(); test4(); system("pause"); return 0; }
标签：内存 改进 -- 解决问题 end pre 结果 数据 use
原文地址：https://www.cnblogs.com/music-liang/p/12053847.html
