力扣算法训练--2.两数相加

2.两数相加题目描述给出两个 非空 的链表用来表示两个非负的整数。其中它们各自的位数是按照 逆序 的方式存储的并且它们的每个节点只能存储 一位 数字。如果我们将这两个数相加起来则会返回一个新的链表来表示它们的和。您可以假设除了数字 0 之外这两个数都不会以 0 开头。示例输入(2 - 4 - 3) (5 - 6 - 4) 输出7 - 0 - 8 原因342 465 807Java题解/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode p l1; //将运算结果保存在l1中 ListNode q l2; int addNum 0; //表示进位数 while(q!null){ //所以以l2遍历为主按值计算 if(p.nextnull q.next!null) // 用于判定链表的结束标志 p.next new ListNode(0); if(q.nextnull p.next!null)// 用于判定链表的结束标志 q.next new ListNode(0); int sumAll addNum p.val q.val; p.val sumAll % 10; //表示本位数保存 addNum sumAll / 10; //用于表示进位数参与下一位数的本位计算 if(p.next null q.next null addNum!0) p.next new ListNode(addNum); //用于结果链的结尾置零表示尾空 p p.next; q q.next; } return l1; } }Java题解2class ListNode { int val; ListNode next; ListNode(int x) { val x; } } class Solution { 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; } }C题解/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ void deeltail(struct ListNode * pre,struct ListNode * r,int carry); struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode * head ,*p,*q,*pre; int carry 0; if(l1 NULL) return l2; if(l2 NULL) return l1; head l1; p head; while(l1 l2){ pre p;//指向p指针的前向指针 p-val l1-val l2-val carry; carry 0; if(p-val 9){ p-val - 10; carry 1; } l1 l1-next; l2 l2-next; p p-next; } if(!l1 !l2){//l1与l2都只有一个数 if(carry){ q (struct ListNode *)malloc(sizeof(struct ListNode)); q-val carry; q-next NULL; pre-next q; } } else if(l2){//l1的长度小于l2,由于head最初是指向l1的特殊处理 deeltail(pre,l2,carry); } else{//l1的长度大于l2 deeltail(pre,l1,carry); } return head; } void deeltail(struct ListNode * pre,struct ListNode * r,int carry){ struct ListNode * q; pre-next r; q pre-next; while(q){ q-val carry; if(q-val9){ q-val - 10; carry 1; }else{ carry 0; } pre q; q q-next; } if(carry){//最后一个节点加上进位大于10需处理多出来的节点val1 q (struct ListNode *)malloc(sizeof(struct ListNode)); q-val carry; q-next NULL; pre-next q; } }C语言题解2/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *new1,*p3; static struct ListNode *result; result(struct ListNode *)malloc(sizeof(struct ListNode)); result-nextNULL;//这里很重要定义了指针以后要把它的next赋成NULL否则会编译出错 int sum0; p3result; while(1) { if(l1) { suml1-val; l1l1-next; } if(l2) { suml2-val; l2l2-next; } p3-valsum%10; sumsum/10; if(l1 || l2 || sum!0)//当l1不为空或l2不为空或sum不为0时循环继续 { new1(struct ListNode *)malloc(sizeof(struct ListNode)); new1-nextNULL;//这里很重要定义了指针以后要把它的next赋成NULL否则会编译出错 p3-nextnew1; p3p3-next; } else break;//否则跳出循环得到结果 } return result; }C题解3/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *p1 l1; struct ListNode *p2 l2; struct ListNode *l3 (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode *p3 l3; int carry 0; int sum 0; while( p1 ! NULL p2 ! NULL) { sum p1-val p2-val carry; if(sum9) { sum%10; carry 1; } else { carry 0; } struct ListNode *p4 (struct ListNode*)malloc(sizeof(struct ListNode)); p4-val sum; p4-next NULL; p3-next p4; p3 p4; p1 p1-next; p2 p2-next; } while(p1!NULL)//p1较长 { sum p1-val carry; if(sum9) { sum%10; carry 1; } else { carry 0; } struct ListNode *p5 (struct ListNode*)malloc(sizeof(struct ListNode)); p5-val sum; p5-next NULL; p3-next p5; p3 p5; p1 p1-next; } while(p2!NULL)//p2较长 { sum p2-val carry; if(sum9) { sum%10; carry 1; } else { carry 0; } struct ListNode *p6 (struct ListNode*)malloc(sizeof(struct ListNode)); p6-val sum; p6-next NULL; p3-next p6; p3 p6; p2 p2-next; } if(carry 1)//完成两条链表的值合并但是最大值位发生进位的处理 { sum carry; struct ListNode *p7 (struct ListNode*)malloc(sizeof(struct ListNode)); p7-val sum; p7-next NULL; p3-next p7; } return l3-next; }