<Leetcode> Add Two Numbers

1. 题目

2. 笔者的解法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    int tmp,carray=0;
    struct ListNode * tmpnode,* headnode,* prenode;
    headnode=calloc(1,sizeof(struct ListNode));
    tmpnode=headnode;
    for(;l1 != NULL || l2 != NULL; ){
        if( l1==NULL && l2 != NULL)
            tmp=l2->val+carray;
        else if(l1 !=NULL && l2 == NULL)
            tmp=l1->val+carray;
        else
            tmp=l1->val+l2->val+carray;

        tmpnode->val=tmp%10;
        carray=tmp/10;
        tmpnode->next=calloc(1,sizeof(struct ListNode));
        prenode=tmpnode;
        tmpnode=tmpnode->next;
        if(l1 != NULL)
            l1=l1->next;
        if(l2 != NULL)
            l2=l2->next;
    }
    if(carray == 1){
        tmpnode->val=1;
        tmpnode->next=NULL;
    }
    else{
        free(tmpnode);
        prenode->next=NULL;
    }
    return headnode;
}

3. 优质解法

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

4. 总结

　　该题笔者和优质解法的思路一样，本题也不涉及其他算法知识，就是简单的操作链表和两数相加的进位处理。且本题目稍作了简化，将数以倒序方式存入链表。否则又要一些代码来处理正序问题。本题算法时间复杂度，假设最大的数的节点数为$n$，则时间复杂度为$O(n)$。