LeetCode的medium题集合(C++实现)六

1 Multiply Strings

Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are non-negative.

该题实际就是利用字符串来解决大数的乘法问题。为了计算方便先将两组数翻转，将字符串转化为整数利用两个循环逐位相乘，将结果保存。然后逐位解决进位问题。

[code]string multiply(string num1, string num2) {
    int flag = 0, len1 = num1.length(), len2 = num2.length();
    reverse(num1.begin(), num1.end());
    reverse(num2.begin(), num2.end());
    vector<int> mid(len1 + len2, 0);
    string result;
    for (int i = 0; i<len1; i++)
    {
        for (int j = 0; j<len2; j++)
        {
            int a = num1[i] - '0';
            int b = num2[j] - '0';
            mid[i + j] += a*b;
        }
    }
    for (int m = 0; m<len1 + len2; m++)
    {
        int c = (mid[m] + flag) % 10;
        flag = (mid[m] + flag) / 10;
        mid[m] = c;
    }
    int n = len1 + len2 - 1;
    while (mid
 == 0)
    {
        n--;   //去掉前面为0的元素
    }
    if (n<0) return result += '0';
    for (; n >= 0; n--)
    {
        result += ('0' + mid
);
    }
    return result;
    }

2 Permutations

Given a collection of numbers, return all possible permutations.For example，[1,2,3] have the following permutations:

[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

前面我们已经解决过nextPermutation问题，即可以求出一组数集的下一个排列，我们可以先将这组数按升序排列，然后循环求解下一个排列直到无解为止，此时可得到这组数全部的排列组合。该方法同时解决了Permutations II.

[code] bool nextPermutation(vector<int>& nums) {
        bool flag=true;
        if (nums.size() < 2) return false;
          int i, k;
          for (i = nums.size() - 2; i >= 0; --i) if (nums[i] < nums[i+1]) break;
          for (k = nums.size() - 1; k > i; --k) if (nums[i] < nums[k]) break;
          if(i<0) flag=false;
          if (i >= 0) swap(nums[i], nums[k]);
          reverse(nums.begin() + i + 1, nums.end());
          return flag;
    }
    vector<vector<int>> permute(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int> > res;
        bool flag=true;
        while(flag)
        {
           res.push_back(nums);
           flag=nextPermutation(nums);
        }
        return res;
    }