【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.

这道题目首先我想到是不能用乘法，那就全部用加法将

然后超时

public class Solution {
public String multiply(String num1, String num2) {
if(num1.equalsIgnoreCase("")||num2.equalsIgnoreCase(""))
return "";
if(num1.equalsIgnoreCase("0")||num2.equalsIgnoreCase("0"))
return "0";
Map<Integer,String> map = new HashMap<Integer,String>();
map.put(0, "0");
map.put(1, num2);
for(int i=2;i<10;i++){
map.put(i, "");
}
char[] ch = new char[num1.length()*num2.length()+1];
Arrays.fill(ch, '0');
int xx=ch.length-1;
for(int i=num1.length()-1;i>=0;i--){
int temp = num1.charAt(i)-'0';
String base="";
if(map.get(temp).equalsIgnoreCase("")){
base=plus(temp,num2);
map.put(temp, base);
}else{
base=map.get(temp);
}
int x=xx;
int y=base.length()-1;
int pre=0;
while(y>=0){
int tt = base.charAt(y)-'0'+ch[x]-'0'+pre;
if(tt>=10){
pre=1;
ch[x]=String.valueOf(tt-10).charAt(0);
}else{
pre=0;
ch[x]=String.valueOf(tt).charAt(0);
}
x--;
y--;
}
if(pre!=0){
if(ch[x]!='0'){
int tm = ch[x]-'0'+pre;
if(tm>=10){
ch[x]='0';
ch[x-1]='1';
}else{
ch[x]=String.valueOf(tm).charAt(0);
}

}else{
ch[x]='1';
}

}
xx--;

}
String re = String.valueOf(ch);
while(re.startsWith("0")){
re = re.substring(1);
}
return re;

}

private String plus(int times,String base){
char[] ch = new char[base.length()+1];
ch[0]='0';
for(int m=base.length()-1;m>=0;m--){
ch[m+1]=base.charAt(m);
}
for(int j=1;j<times;j++){
int pre=0;
int i=base.length()-1;
int k=i+1;
while(i>=0){
int temp = base.charAt(i)-'0'+ch[k]-'0'+pre;
if(temp>=10){
pre=1;
ch[k]=String.valueOf(temp-10).charAt(0);
}else{
pre=0;
ch[k]=String.valueOf(temp).charAt(0);
}
i--;
k--;
}
if(pre!=0){
if(ch[k]!='0'){
int tm = ch[k]-'0'+pre;
if(tm>=10){
ch[k]='0';
ch[k-1]='1';
}else{
ch[k]=String.valueOf(tm).charAt(0);
}

}else{
ch[k]='1';
}
}

}
if(ch[0]=='0'){
return String.valueOf(ch, 1, ch.length-1);

}else{
return String.valueOf(ch);
}

}
}

后面再网上看到一种巧妙的解法。首先将每个位的数求出来，然后再遍历是不是需要进位

public class NSolution {
public String multiply(String num1, String num2) {
String n1 = new StringBuilder(num1).reverse().toString();
String n2 = new StringBuilder(num2).reverse().toString();
int[] re = new int[n1.length()+n2.length()];
　　　　//把i j 设为 0 则 i+j就是新的数的第i+j位，很巧妙

for(int i=0;i<n1.length();i++){
for(int j=0;j<n2.length();j++)
re[i+j] += (n1.charAt(i)-'0') * (n2.charAt(j)-'0');
}
StringBuilder sb = new StringBuilder();
for(int k=0;k<re.length;k++){
int low = re[k]%10;
int high = re[k]/10;
if(k+1<re.length)
re[k+1]+=high;
sb.insert(0, low);
}
while(sb.charAt(0)=='0' && sb.length()>1){
sb.deleteCharAt(0);
}
return sb.toString();
}
}