标签：两个 num1 num2 int pos length 字符串 sb 乘法
见leetcode43
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
- The length of both
num1 and num2 is < 110.
- Both
num1 and num2 contains only digits 0-9.
- Both
num1 and num2 does not contain any leading zero.
- 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]
class Solution {
public String multiply(String num1, String num2) {
int m = num1.length();
int n = num2.length();
int[] pos = new int[m + n];
for (int i = m - 1; i >= 0; i--) {
for (int j = n - 1; j >= 0; j--) {
int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
int p1 = i + j;
int p2 = i + j + 1;
int sum = mul + 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))
sb.append(p);
return sb.length() == 0 ? "0" : sb.toString();
}
}
标签：两个,
num1,
num2,
int,
pos,
length,
字符串,
sb,
乘法
From： https://www.cnblogs.com/MarkLeeBYR/p/16812626.html