43. Multiply Strings(大数相乘)
2016-03-03 22:18
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问题描述:
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.
问题求解:
Space cost O(n)
Time cost O(n)
代码如下:
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.
问题求解:
Space cost O(n)
Time cost O(n)
代码如下:
class Solution { public: string multiply(string num1, string num2) { int n1=num1.size(); int n2=num2.size(); int n=n1+n2; if(num1=="0" || num2=="0" || n1==0 || n2==0) return "0"; vector<int> a(n,0);//乘积数组 string str="";//返回的字符串结果 for(int i=n1-1;i>=0;i--) {//从后往前是位数升高的方向 for(int j=n2-1;j>=0;j--) {//(1)得到乘积数组各个位上的数(还未处理进位) a[i+j+1] += (num1[i]-'0')*(num2[j]-'0'); } } //如a[i]=12,a[i-1]=3,处理进位:a[i-1]=3 + 12/10=4;a[i]=12%10=2; for(int i=n-1;i>0;i--) {//(2)从低位到高位依次处理进位.高位a[i-1],低位a[i] a[i-1] += a[i]/10;//高位=高位初始乘积数+进位 a[i] %= 10; } int i=0; if(a[i]==0) {//(3)若最高位是0,则处理掉 i++; } while(i<n) {//(4)将数组中的结果依次转换为字符加到str中 str += a[i++]+'0';//注意i++!!! } return str; } };