Leetcode | 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.

还是不要想得太复杂。一开始还想着要4位4位二分去做。其实只要一位一位地乘，小心地维护好进位就好。

 1 class Solution {

 2 public:

 3     string multiply(string num1, string num2) {

 4         if (num1.empty()) return "";

 5         if (num2.empty()) return "";

 6         int m1, m2, carry, val;

 7         

 8         reverse(num1.begin(), num1.end());

 9         reverse(num2.begin(), num2.end());

10         string ret(num1.length() + num2.length(), '0');

11         for (int i = 0; i < num2.length(); ++i) {

12             m2 = num2[i] - '0';

13             carry = 0;

14             for (int j = 0; j < num1.length(); ++j) {

15                 m1 = num1[j] - '0';

16                 val = m2 * m1 + carry + ret[i + j] - '0';

17                 ret[i + j] = val % 10 + '0';

18                 carry = val / 10;

19             }

20             ret[i + num1.length()] += carry;

21         }

22 

23         string r = "0";

24         for (int i = ret.length() - 1; i >= 0; --i) {

25             if (ret[i] != '0') {

26                 r = ret.substr(0, i + 1);

27                 break;

28             }

29         }

30         reverse(r.begin(), r.end());

31         return r;

32     }

33 };

reverse主要是便于取下标。

第10行，注意结果最大长度就是两个串的长度和；

第16行，注意每个位置需要加上ret[i+j]-'0'; 第17行，直接等于新的余数；

第20行，记得要把最后的进位保存下来；

第23行，如果从后往前扫一直没找到非零的数，那么证明结果为0，所以默认值设为'0'；

感觉第三次写比较长， 但是代码清晰许多。

 1 class Solution {

 2 public:

 3     string mul(string &num1, int n) {

 4         if (n == 0 || num1 == "0") return "0";

 5         if (n == 1) return num1;

 6         string ans = "";

 7         int carry = 0, v;

 8         for (int i = num1.length() - 1; i >= 0; --i) {

 9             v = carry + (num1[i] - '0') * n;

10             ans.push_back((v % 10) + '0');

11             carry = v / 10;

12         }

13         if (carry > 0) ans.push_back(carry + '0');

14         reverse(ans.begin(), ans.end());

15         return ans;

16     }

17     

18     void add(string &ans, string &num) {

19         int n1 = ans.length(), n2 = num.length(), v = 0, carry = 0;

20         string tmp = "";

21         for (int i = n1 - 1, j = n2 - 1; i >= 0 || j >= 0; --i, --j) {

22             v = carry;

23             if (i >= 0) v += (ans[i] - '0');

24             if (j >= 0) v += (num[j] - '0');

25             tmp.push_back((v % 10) + '0');

26             carry = v / 10;

27         }

28         if (carry > 0) tmp.push_back(carry + '0');

29         reverse(tmp.begin(), tmp.end());

30         ans = tmp;

31     }

32     

33     string multiply(string num1, string num2) {

34          if (num1.empty() || num2.empty()) return "";

35          string ans = "0";

36          for (int i = num2.length() - 1; i >= 0; --i) {

37              string sum = mul(num1, num2[i] - '0');

38              if (num2.length() - i - 1 > 0 && sum != "0") sum.append(num2.length() - i - 1, '0');

39              add(ans, sum);

40          }

41          return ans;

42     }

43 };