LeetCode Unique Paths 动态规划与大数

来源：互联网发布：斗牛怎么玩法算法编辑：程序博客网时间：2024/06/06 18:32

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

由于只能往下或者往右走，因此(i, j)只会由(i - 1, j)或者(i, j - 1)到达。

假设，到达(i - 1, j)有f[i - 1, j]种走法，到达(i, j - 1)有f[i, j - 1]种走法，那么到达(i, j)有f[i, j] = f[i - 1, j] + f[i, j - 1]中走法。

/* * kl.cpp * *  Created on: 2014年12月27日 *      Author: judyge */#include<cstring>#include<iomanip>#include<algorithm>#include<cstdlib>#include<cstdio>#include<iostream>#include<vector>#include<stdio.h>#include<time.h>#include<windows.h>#define MAXN 9999#define MAXSIZE 1000#define DLEN 4using namespace std;class BigNum { private:     int a[500];    //可以控制大数的位数     int len;       //大数长度 public:     BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数     BigNum(const int);     friend ostream& operator<<(ostream&,  BigNum&);     BigNum operator+(const BigNum &) const;     void print(); }; BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数 {     int c,d = b;     len = 0;     memset(a,0,sizeof(a));     while(d > MAXN)     {         c = d - (d / (MAXN + 1)) * (MAXN + 1);         d = d / (MAXN + 1);         a[len++] = c;     }     a[len++] = d; } ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符 {     int i;     cout << b.a[b.len - 1];     for(i = b.len - 2 ; i >= 0 ; i--)     {         cout.width(DLEN);         cout.fill('0');         cout << b.a[i];     }     return out; } BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算 {     BigNum t(*this);     int i,big;      //位数     big = T.len > len ? T.len : len;     for(i = 0 ; i < big ; i++)     {         t.a[i] +=T.a[i];         if(t.a[i] > MAXN)         {             t.a[i + 1]++;             t.a[i] -=MAXN+1;         }     }     if(t.a[big] != 0)         t.len = big + 1;     else         t.len = big;     return t; } void BigNum::print()    //输出大数 {     int i;     cout << a[len - 1];     for(i = len - 2 ; i >= 0 ; i--)     {         cout.width(DLEN);         cout.fill('0');         cout << a[i];     }     cout << endl; }BigNum uniquePaths(int m, int n) {        vector<vector<BigNum> > v(m, vector<BigNum>(n, 1));        for(int i=1; i<m; ++i){            for(int j=1; j<n; ++j){                v[i][j]=v[i-1][j]+v[i][j-1];            }        }        return v[m-1][n-1];    } int main(){clock_t start,finish;double time;start=clock();uniquePaths(1000,500).print();finish=clock();time=(double)((finish-start)/CLOCKS_PER_SEC);printf("start:%ld\t\tfinish:%ld\tfinish-start:%ld\truntime:%f\n",start,finish,finish-start,time);return 0; }

运行结果

21794643772475632607161961769677171260214971666986831475298702747021969894224312122274198675876219732944354758456178809790971668241976102754811035273030233861576123498221326050261052311668799203434574640707827487765370771640762604652965442259548989227841052691939812694812193778630470756833909910697283456008280175193445777429087108659795300989641974873048618112352532398954469345334076036542958624176451332960000start:10finish:1505finish-start:1495runtime:1.000000

动态规划效果再次显示出来.1s搞定.不过360检测内存消耗非常大.

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