LeetCode OJ--Unique Paths *

https://oj.leetcode.com/problems/unique-paths/

首先，转换成一个排列组合问题，计算组合数C(m+n-2) (m-1),请自动想象成上下标。

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 1 && n == 1)
            return 1;
        int sum1 = 1;
        int sum2 = 1;
        for(int i = 1; i <= m-1; i++)
            sum1 *= i;
        for(int j = m+n-2; j >= n; j--)
            sum2 = sum2*j;
        return sum2/sum1;
    }
};

runtime error,当测试数据是36,7的时候，也就是说，这个数太大了，已经算不了了。

于是参考了discuss

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 1 && n == 1)
            return 1;
        int sum1 = 1;
        int sum2 = 1;
        //exchange;
        int temp;
        if(m<n)
        {
            temp = n; n = m; m = temp;
        }
        int p,q;
        int commonFactor;
        for(int i = 1; i<= n-1; i++)
        {
            p = i;
            q = i+m-1;
            commonFactor = gcd(p,q);
            sum1 = sum1 * (p/commonFactor);
            sum2 = sum2 * (q/commonFactor);
            commonFactor = gcd(sum1,sum2);
            sum1 = sum1/commonFactor;
            sum2 = sum2/commonFactor;
        }

        return sum2/sum1;
    }

    int gcd(int a, int b) 
    {
        while(b)
        {
            int c = a%b;
            a = b;
            b = c;
        }
        return a;
    }
};

输入58,61的时候wa，因为结果得了个负数，明显又溢出了。

于是：

#include <iostream>
using namespace std;

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 1 && n == 1)
            return 1;
        long long sum1 = 1;
        long long sum2 = 1;
        //exchange;
        int temp;
        if(m<n)
        {
            temp = n; n = m; m = temp;
        }
        int p,q;
        int commonFactor;
        for(int i = 1; i<= n-1; i++)
        {
            p = i;
            q = i+m-1;
            commonFactor = gcd(p,q);
            sum1 = sum1 * (p/commonFactor);
            sum2 = sum2 * (q/commonFactor);
            commonFactor = gcd(sum1,sum2);
            sum1 = sum1/commonFactor;
            sum2 = sum2/commonFactor;
        }

        return sum2/sum1;
    }

    int gcd(long long a, long long b) 
    {
        while(b)
        {
            int c = a%b;
            a = b;
            b = c;
        }
        return a;
    }
};

int main()
{
    Solution myS;
    cout<<myS.uniquePaths(51,9);
    return 0;
}

中间过程中使用了long long 类型。

记住求公约数的算法。