两个字符串的乘法

见leetcode43

 

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

https://discuss.leetcode.com/topic/30508/easiest-java-solution-with-graph-explanation

//num1的第i位和num2的第j位相乘，结果存在结果字符串的[i+j, i+j+1]

 

public String multiply(String num1, String num2) {
  int m = num1.length();
  int n = num2.length();
  int[] pos = new int[m + n]; // m位数和n位数相乘，结果最多的位数可能是m+n
  for (int i = m - 1; i >= 0; i--) { // 个位在字符串的最后一位，所以从length-1开始遍历。外层循环表示乘数
    for (int j = n - 1; j >= 0; j--) { // 内层循环表示被乘数
      int p1 = i + j;
      int p2 = i + j + 1;
      int sum = (num1.charAt(i) - '0') * (num2.charAt(j) - '0') + pos[p2]; //此时的pos[p2]就是进位值

      pos[p1] += sum / 10;
      pos[p2] = sum % 10;
    }
  }

  StringBuilder sb = new StringBuilder();
  for (int p : pos) {
    if (sb.length() == 0 && p == 0) {
      continue;
    }
    sb.append(p);
  }
  return sb.length() == 0 ? "0" : sb.toString();
}