问题
给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
示例:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出:7 -> 0 -> 8 原因:342 + 465 = 807
代码实现
#include <vector> #include <map> #include <iostream> #include <math.h> /** * 给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。 你可以假设除了数字 0 之外,这两个数字都不会以零开头。 示例: 输入:(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出:7 -> 0 -> 8 原因:342 + 465 = 807 */ using namespace std; struct ListNode{ int val; ListNode *next; ListNode(int x) : val(x), next(NULL){} }; template<class T> int length(T& arr) { return sizeof(arr)/ sizeof(arr[0]); } class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { uint64_t carry = 0, sum = 0; ListNode prehead(0), *p = &prehead; while (l1 || l2) { sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry; sum = carry / 10; p->next = new ListNode(sum % 10); l1 = l1 ? l1->next : l1; l2 = l2 ? l2->next : l2; p = p->next; } return prehead.next; } }; int main(int argc, char** argv){ int a[] = {2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,9}; ListNode* tmp = new ListNode(0); ListNode* ptr1 = tmp; for (int i = 0; i < length(a); i++) { ptr1->next = new ListNode(a[i]); ptr1 = ptr1->next; } ptr1 = tmp->next; delete tmp; tmp = new ListNode(0); int b[] = {5,6,4,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,2,4,3,9,9,9,9}; ListNode* ptr2 = tmp; for (int j = 0; j < length(b); j ++) { ptr2->next = new ListNode(b[j]); ptr2 = ptr2->next; } ptr2 = tmp->next; delete tmp; Solution* solution = new Solution(); solution->addTwoNumbers(ptr1, ptr2); delete solution; return 0; }
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