题解 | #牛群的能量值#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 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
 *
 *
 * @param l1 ListNode类
 * @param l2 ListNode类
 * @return ListNode类
 */
#include <stdio.h>
int getLength(struct ListNode* l1) {
    int count = 0;
    while (l1) {
        count++;
        l1 = l1->next;
    }
    return count;
}
struct ListNode* addEnergyValues(struct ListNode* l1, struct ListNode* l2 ) {
    // write code here
    int len1 = getLength(l1);
    int len2 = getLength(l2);
    int cin = 0;
    if (len1 >= len2) {
        struct ListNode* res = l1;
        struct ListNode* res1 = l1;
        while (l2) {
            int num = (l1->val + l2->val + cin) % 10;
            cin = (l1->val + l2->val + cin) / 10;
            l1->val = num;
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1) {
            int num = (l1->val + cin) % 10;
            cin = (l1->val + cin) / 10;
            l1->val = num;
            l1 = l1->next;
        }
        if (cin != 0) {
            struct ListNode* temp = malloc(sizeof(struct ListNode));
            temp->val = cin;
            temp->next = NULL;
            while(res1){
                if(res1->next== NULL){
                    res1->next=temp;
                    res1=temp;
                    break;
                }
                res1=res1->next;
            }
        }
        return res;
    } else {
        struct ListNode* res = l2;
        struct ListNode* res1 = l2;
        while (l1) {
            int num = (l1->val + l2->val + cin) % 10;
            cin = (l1->val + l2->val + cin) / 10;
            l2->val = num;
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l2) {
            int num = (l2->val + cin) % 10;
            cin = (l2->val + cin) / 10;
            l2->val = num;
            l2 = l2->next;
        }
        if (cin != 0) {
            struct ListNode* temp = malloc(sizeof(struct ListNode));
            temp->val = cin;
            temp->next = NULL;
            while(res1){
                if(res1->next== NULL){
                    res1->next=temp;
                    res1=temp;
                    break;
                }
                res1=res1->next;
            }
        }
        return res;
    }
}

优先判断链表l1，l2的长度，在长的链表进行相加操作，若计算到最后进位不为0，则进行插入操作