[leetcode]43. Multiply Strings高精度乘法
Given two non-negative integers num1
and num2
represented as strings, return the product of num1
and num2
, also represented as a string.
Example 1:
Input: num1 = "2", num2 = "3"
Output: "6"
Example 2:
Input: num1 = "123", num2 = "456"
Output: "56088"
Note:
- The length of both
num1
andnum2
is < 110. - Both
num1
andnum2
contain only digits0-9
. - Both
num1
andnum2
do not contain any leading zero, except the number 0 itself. - You must not use any built-in BigInteger library or convert the inputs to integer directly.
题意:
高精度乘法。
Solution1: Math
Do the simulation like how computer will do multiplication operation
1一个char对应1个digit,
digit相乘后,注意其乘积结果可能需要进位
code
/*
Time: O(n^2). We use nested 2 for loop
Space: O(n). We use int[] to save intermedia infor
*/ class Solution {
public String multiply(String num1, String num2) {
if(num1.length()==0 || num2.length()==0) return "0";
int len1 = num1.length();
int len2 = num2.length();
int [] result = new int [len1+len2]; for(int i = len1-1; i>=0; i--){
for(int j = len2-1; j>=0;j--){
int mul = (num1.charAt(i)-'0')*(num2.charAt(j)-'0');
int idx = i+j+1;
// 可能会进位
int carryIdx = i+j;
mul = mul + result[idx];
result[idx] = mul % 10;
result[carryIdx] = result[carryIdx] + mul/10;
}
}
StringBuilder sb = new StringBuilder();
for(int res: result){
if(sb.length()!=0 || res!=0) sb.append(res);
}
return (sb.length() == 0)? "0" : sb.toString();
}
}
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