【LeetCode】2. Add Two Numbers

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本文連結：https://blog.csdn.net/shiliang97/article/details/101436465

題目描述

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.           

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我太難了，感覺自己C語言文法上咋都有這麼多問題呢~~~~各種指針異常，以前刷pat都沒遇到過。現在問題全都暴露出來了

先研究别人的優秀的代碼，快速的過一遍吧，回過頭來再看。反正資料結構啥的我都還沒有學呢~~~

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    int flag = 0, nodeSum = 0, freeflag = 0;
    struct ListNode *ret,*now,*high,*freeBegin,*freeEnd;
    for(ret = l1, now = l1, high = l2, freeBegin = l2;l1  || l2  || flag ;) {  
        if (l1 == NULL && l2 == NULL) {
              now->next = high;
              now = now->next;
              freeBegin = now->next;
        }
        nodeSum = ( l1  ? l1->val : 0 ) + (l2 ? l2->val : 0) + flag;  
        now->val = nodeSum % 10;    
        flag =  nodeSum / 10; 
        l1 ? l1 = (l1->next ? l1->next : NULL) : NULL;
        if(l1 == NULL && 0 == freeflag && l2) {
            freeEnd = l2;
            freeflag = 1;
        }
        l2 ? l2 = (l2->next ? l2->next : NULL) : NULL;
        now->next = ( l1 ? l1 : (l2 ?  l2 : NULL) );
        now->next ? now = now->next : NULL;
    }
    l2 = freeBegin;
    freeEnd->next = NULL;
    return ret;
}           

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