leetcode Add Two Numbers
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思路:
1,新建第三个个链表,用来存储得到的和
2 ,判断给的两个字符串的长度,确定两个数的位数
3,从低位开始相加,如果超过十,则向高位进1;不足十的话直接保存
4,当数位低的链表加完后,如果没有进位,则所求数的高位数和较长的数组的高位数一样
class Solution {public: ListNode addTwoNumbers(ListNode *l1, ListNode *l2) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. int carry = 0; ListNode *head; ListNode *pre; ListNode *p1 = l1; ListNode *p2 = l2; while(p1&&p2){ p1=p1->next; p2=p2->next; } if(p2){ p2=l1; p1=l2; } else{ p1=l1; p2=l2; } head = p1; while(p1){ p1->val =p1->val+carry+(p2? p2->val:0); carry = (p1->val)/10; p1->val=(p1->val)%10; pre = p1; p1=p1->next; if(p2){ p2=p2->next; } } if(carry){ pre->next=new ListNode(1); } return head; }};
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