大整数相加

思路

• 将大整数的字符串数组从后向前逐位相加
• 用dif表示两个数组长度的差，takeOver 表示每一位计算后的进位，长度为最大长度+1的result字符数组存储相加后的结果
• 先计算后dif位的和；
• 如果dif>0，再将剩下的高位与地位的进位相加，拷贝到result数组的前dif位，得到结果；
• 如果dif==0, 直接将进位放到result[0]的位置，得到结果；

C++实现代码如下：

#include 

using namespace std;

char * numberAdd(char * number1, char * number2){
	int len1 = strlen(number1);
	int len2 = strlen(number2);
	char * min_len_number;
	char * max_len_number;
	int min_len,max_len;
	int dif;
	if(len1>=len2){
		min_len = len2;
		min_len_number = number2;
		max_len = len1;
		max_len_number = number1;
	}
	else{
		min_len = len1;
		min_len_number = number1;
		max_len = len2;
		max_len_number = number2;
	}
	dif = max_len-min_len;
	char * result = new char[max_len+2];
	memset(result,'0',max_len+1);
	result[max_len+1] = '\0';
	int takeOver = 0;
	for(int i=max_len-1;i>=dif;i--){
		int sum = (max_len_number[i]-'0')+(min_len_number[i-dif]-'0')+takeOver;
		takeOver = 0;
		if(sum>=10){
			takeOver = 1;
			sum -= 10;
			result[i+1] = sum + '0';
		}
		else{
			result[i+1] = sum+'0';
		}
		cout<0){
		for(int i=dif-1;i>=0;i--){
			int sum = max_len_number[i]-'0'+takeOver;
			takeOver = 0;
			if(sum>=10){
				if(i==0){
					result[i] = '1';
				}
				else{
					takeOver = 1;
					sum -= 10;
					result[i+1] = sum + '0'; 	
				}
			}
			else{
				result[i+1] = sum + '0';
			}
		}
	}
	else{
		result[0] = takeOver+'0';
	}
//	cout<