给定两个用链表表示的整数,每一个节点包含一个数位。面试
这些数位是反向存放的,也就是个位排在链表首部。算法
编写函数对这两个整数求和,并用链表形式返回结果。函数
试题连接:https://leetcode-cn.com/problems/sum-lists-lcci/测试
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode pCurrent = l1; //遍历 int index = 0; int num1 = 0; while(pCurrent != null) { int num = pCurrent.val; //个、10、百依次 num1 += num * Math.pow(10,index); index++; pCurrent = pCurrent.next; } //重复 pCurrent = l2; index = 0; int num2 = 0; while(pCurrent != null) { int num = pCurrent.val; //个、10、百依次 num2 += num * Math.pow(10,index); index++; pCurrent = pCurrent.next; } //进行相加 int sum = num1 + num2; // System.out.println(sum); //将数字拆分,转换为链表 912 2 91 1 9 9 0 ListNode listNode = new ListNode(sum % 10); sum = sum / 10; pCurrent = listNode; while (sum != 0) { int x = sum % 10; pCurrent.next = new ListNode(x); pCurrent = pCurrent.next; sum = sum / 10; } return listNode; }
测试结果:3d
由于测试数值较大,int类型没法正确的进行保存,故出现了错误。将int类型改成long类型,在进行测试。code
发现所测试的数据远远比咱们想一想的大,所以咱们得另辟蹊径。blog
考虑到该算法有超大规模的测试数据,咱们引入了BigInteger这个类,来进行测试。ci
import java.math.BigInteger; class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode pCurrent = l1; //遍历 String num1 = ""; while(pCurrent != null) { int num = pCurrent.val; //个、10、百依次 num1 += num; pCurrent = pCurrent.next; } //重复 pCurrent = l2; String num2 = ""; while(pCurrent != null) { int num = pCurrent.val; //个、10、百依次 num2 += num; pCurrent = pCurrent.next; } //逆序遍历,进行相加 String newNum1 = ""; for(int i = 0;i < num1.length();i++) { newNum1 += num1.charAt(num1.length() - 1 - i); } String newNum2 = ""; for(int i = 0;i < num2.length();i++) { newNum2 += num2.charAt(num2.length() - 1 - i); } // System.out.println(newNum1); // System.out.println(newNum2); BigInteger bigInteger1 = new BigInteger(newNum1); BigInteger bigInteger2 = new BigInteger(newNum2); BigInteger sum = bigInteger1.add(bigInteger2); String sumStr = sum.toString(); ListNode saveNode = new ListNode(sumStr.charAt(sumStr.length() - 1) - 48); pCurrent = saveNode; for(int i = 1;i < sumStr.length();i++) { pCurrent.next = new ListNode(sumStr.charAt(sumStr.length() - 1 - i) - 48); pCurrent = pCurrent.next; } return saveNode; } }
测试结果:leetcode
能够看到,该算法的用时较长,且若是改写成C语言代码时,咱们并无BigInteger这个类。
对此,咱们像将链表拆分,引入对位相加的计算策略。
ListNode list = new ListNode(-1); //定义输出链表 ListNode p = list; int num = 0; //进位数字 int x = 0; //记录l1链表的值 int y = 0; //记录12链表的值 //遍历两个链表 while(l1 != null ||l2 != null) { x = l1 == null ? 0 : l1.val; y = l2 == null ? 0 : l2.val; int sum = x + y + num; if(sum < 10) { p.next = new ListNode(sum); num = 0; }else { p.next = new ListNode(sum % 10); num = sum / 10; } if(l1 != null) l1 = l1.next; if(l2 != null) l2 = l2.next; p = p.next; } if (num != 0) p.next = new ListNode(num); return list.next;
测试结果:
能够看到,对于该算法,在java中的运行效率仍是能够的。
struct ListNode* list = (struct ListNode*)malloc(sizeof(struct ListNode)); list->val = -1; //定义输出链表 list->next = NULL; struct ListNode* p = list; int num = 0; //进位数字 int x = 0; //记录l1链表的值 int y = 0; //记录12链表的值 //遍历两个链表 while(l1 != NULL ||l2 != NULL) { x = l1 == NULL ? 0 : l1->val; y = l2 == NULL ? 0 : l2->val; int sum = x + y + num; if(sum < 10) { struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode)); temp->val = sum; temp->next = NULL; p->next = temp; num = 0; }else { struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode)); temp->val = sum % 10; temp->next = NULL; p->next = temp; num = sum / 10; } if(l1 != NULL) l1 = l1->next; if(l2 != NULL) l2 = l2->next; p = p->next; } if (num != 0) { struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode)); temp->val = num; temp->next = NULL; p->next = temp; } return list->next;
测试结果:
能够看到,在C语言中,该算法的计算效率偏低。