LeetCode 445. Add Two Numbers II(链表求和)
题意:两个非空链表求和,这两个链表所表示的数字没有前导零,要求不能修改原链表,如反转链表。
分析:用stack分别存两个链表的数字,然后从低位开始边求和边重新构造链表。
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { stack<int> s1, s2; while(l1){ s1.push(l1 -> val); l1 = l1 -> next; } while(l2){ s2.push(l2 -> val); l2 = l2 -> next; } ListNode *cur = new ListNode(0); while(!s1.empty() || !s2.empty()){ if(!s1.empty()){ cur -> val += s1.top(); s1.pop(); } if(!s2.empty()){ cur -> val += s2.top(); s2.pop(); } ListNode* pre = new ListNode(cur -> val / 10); cur -> val %= 10; pre -> next = cur; cur = pre; } return (cur -> val == 0) ? cur -> next : cur; } };