题解 | #大数加法#

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 计算两个数之和
 * @param s string字符串 表示第一个整数
 * @param t string字符串 表示第二个整数
 * @return string字符串
 */
#include <stdlib.h>
#include <string.h>
char* solve(char* s, char* t ) {
    // write code here
    int sLen = strlen(s);
    int tLen = strlen(t);
    int retLen = 0;

    if(0 == sLen){
        return t;
    }
    if(0 == tLen){
        return s;
    }

    if(sLen >= tLen){
        retLen = sLen + 2;
    }else{
        retLen = tLen + 2;
    }
    //char *result = (char*)malloc(retLen);
    char *result = (char*)calloc(retLen, sizeof(char));
    int sPos = sLen - 1;
    int tPos = tLen - 1;
    int retPos = 0;
    int jinWei = 0;
    int sum = 0;
    int i, j;

    while(sPos >= 0 || tPos >= 0 || jinWei > 0){
        int sInt = sPos >= 0 ? s[sPos--] - '0' : 0;
        int tInt = tPos >= 0 ? t[tPos--] - '0' : 0;
        sum = sInt + tInt + jinWei;
        if(sum >= 10){
            result[retPos++] = sum % 10 + '0';
            jinWei = 1;
        }else{
            result[retPos++] = sum + '0';
            jinWei = 0;
        }
    }
    //result[retPos] = '\0';
    for(int i = 0, j = retPos - 1; i<j; i++, j--){
        char t = result[i];
        result[i] = result[j];
        result[j] = t;
    }
    
    return result;
}