思路来源：： https://blog.csdn.net/lichong_87/article/details/6860329
/**
* @date 2018/6/22
* @description
*/
public class BigNumUtil { /**
* 大数相乘
* @param a
* @param b
* @return
*/
public static String multi(String a,String b){ //1.判断相乘之后的符号
char signA = a.charAt(0);
char signB = b.charAt(0);
boolean isPositive = true;
if(signA == '+' || signA == '-'){
if(signA == '+'){
isPositive = true;
}else{
isPositive = false;
}
a = a.substring(1);
} if(signB == '+' || signB == '-'){
if(signB == '+' && isPositive){
isPositive = true;
}else if(signB == '-' && !isPositive){
isPositive = true;
}else{
isPositive = false;
}
b = b.substring(1);
} //2.反转数字，使用倒序
StringBuilder sb = new StringBuilder(a);
char[] numA = sb.reverse().toString().toCharArray();
sb = new StringBuilder(b);
char[] numB = sb.reverse().toString().toCharArray(); //3.乘法规律:
// 对 a * b 而言，a 的第 i 位和 b 的第 j 位相乘的结果会放在 result[i+j]中
// 两个数（a,b）相乘最后的结果位数一定小于或等于 a.len + b.len ;
int[] result = new int[numA.length + numB.length];
for(int i = 0 ; i < numA.length ; i++){
for(int j = 0 ; j < numB.length ; j++){
result[i+j] += (numA[i] - '0') * (numB[j] - '0');
}
} //4.实现进位运算
for(int i = 0 ; i < result.length ; i++){
if(result[i] > 10){
result[i+1] += result[i] / 10;
result[i] = result[i] % 10 ;
}
} //5.输出结果
StringBuilder re = isPositive ? new StringBuilder() : new StringBuilder("-");
boolean bool = true;
for(int i = result.length-1 ; i >=0 ; i--){
if(result[i] == 0 && bool){
continue;
}else{
bool = false;
}
re.append(result[i]);
} return re.toString();
} /**
* 大数相减
*/
public static String sub(String a ,String b){ char[] numA = new StringBuilder(a).reverse().toString().toCharArray();
char[] numB = new StringBuilder(b).reverse().toString().toCharArray(); //2.判断符号位
boolean isPositive = true;
if(numA.length < numB.length){
isPositive = false;
}else if(numA.length == numB.length){
int i = numA.length - 1 ;
while (i > 0 && numA[i] == numB[i]){
i--;
}
if(numA[i] < numB[i]){
isPositive = false;
}
} //3.开始计算
int maxLen = numA.length > numB.length ? numA.length : numB.length;
int result[] = new int[maxLen];
for(int i = 0 ; i < maxLen ; i++){
int intA = i < numA.length ?(numA[i] - '0') : 0;
int intB = i < numB.length ?(numB[i] - '0') : 0;
if(isPositive){
result[i] = intA - intB;
}else{
result[i] = intB - intA;
}
} //4.进位装换
for(int i = 0 ; i < result.length ; i++){
if(result[i] < 0){
result[i+1] -=1;
result[i] += 10;
}
} //5.输出结果
StringBuilder re = isPositive ? new StringBuilder() : new StringBuilder("-");
boolean bool = true;
for(int i = result.length - 1 ; i >= 0 ; i--){
if(result[i] == 0 && bool){
continue;
}else{
bool = false;
}
re.append(result[i]);
} return re.toString();
} /**
* 大数相加
* @param a
* @param b
* @return
*/
public static String add(String a,String b){ //1.判断相加之后的符号
char signA = a.charAt(0);
char signB = b.charAt(0); if(signA == '-' && signB == '-'){//A B 都是负的
return "-" + add(a.substring(1),b.substring(1));
}else if((signA == '-' && signB != '-')){//A 是负的 B是正的
return sub(b.substring(1),a.substring(1));
}else if(signA != '-' && signB == '-'){//A是正的 B是负的
return sub(a.substring(1),b.substring(1));
} if(signA == '-' || signA == '+'){
a = a.substring(1);
}
if(signB == '-' || signB == '+'){
b = b.substring(1);
} char[] numA = new StringBuilder(a).reverse().toString().toCharArray();
char[] numB = new StringBuilder(b).reverse().toString().toCharArray(); //2.开始计算
int[] result = new int[numA.length + 1];
for(int i = 0 ; i < result.length;i++){
int intA = i < numA.length ? numA[i] - '0' : 0;
int intB = i < numB.length ? numB[i] - '0' : 0;
result[i] = intA + intB;
}
//3.进位转换
for(int i = 0 ; i < result.length ; i++){
if(result[i] > 10){
result[i+1] += result[i] /10;
result[i] = result[i] % 10 ;
}
}
//4.输出结果
StringBuilder re = new StringBuilder();
boolean bool = true;
for(int i = result.length -1 ; i >=0 ; i--){
if(result[i] == 0 && bool){
continue;
}else{
bool = false;
}
re.append(result[i]);
}
return re.toString();
} }