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。

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
   
     mid(len1 + len2, 0); string result; for (int i = 0; i
    
     = 0; n--) { result += ('0' + mid[n]); } 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.

 bool nextPermutation(vector
   
    & 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
    
     > permute(vector
     
      & nums) { sort(nums.begin(),nums.end()); vector
      
        > res; bool flag=true; while(flag) { res.push_back(nums); flag=nextPermutation(nums); } return res; }