2. Add Two Numbers
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Mean:
给你两个数字链表，让你将两个链表相加，结果保存在一个新链表中.
analyse:
最基本的链表操作.
做链表题时只需注意：先分配(new ListNode(val)),再h=h->next.也就是不要将指针指向空指针.
Time complexity: O(N)
view code
/** * ----------------------------------------------------------------- * Copyright (c) 2016 crazyacking.All rights reserved. * ----------------------------------------------------------------- * Author: crazyacking * Date : 2016-01-29-16.16 */ #include <queue> #include <cstdio> #include <set> #include <string> #include <stack> #include <cmath> #include <climits> #include <map> #include <cstdlib> #include <iostream> #include <vector> #include <algorithm> #include <cstring> using namespace std; typedef long long( LL); typedef unsigned long long( ULL); const double eps( 1e-8); // 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) { int jinwei = 0; ListNode * h1 = l1; ListNode * h2 = l2; ListNode * ans , * ret; bool isFirst = true; while( h1 && h2) { int val = h1 -> val + h2 -> val + jinwei; int now = val % 10; jinwei = val / 10; if( isFirst) { ans = new ListNode( now); ret = ans; isFirst = false; } else { ans -> next = new ListNode( now); ans = ans -> next; } h1 = h1 -> next; h2 = h2 -> next; } while( h1) { int val = h1 -> val + jinwei; int now = val % 10; jinwei = val / 10; if( isFirst) { ans = new ListNode( now); ret = ans; isFirst = false; } else { ans -> next = new ListNode( now); ans = ans -> next; } h1 = h1 -> next; } while( h2) { int val = h2 -> val + jinwei; int now = val % 10; jinwei = val / 10; if( isFirst) { ans = new ListNode( now); ret = ans; isFirst = false; } else { ans -> next = new ListNode( now); ans = ans -> next; } h2 = h2 -> next; } while( jinwei) { ans -> next = new ListNode( jinwei % 10); jinwei /= 10; ans = ans -> next; } return ret; } }; int main() { int n1 , n2; while( cin >> n1 >> n2) { ListNode * h1 , * head1; ListNode * h2 , * head2; int tmp; for( int i = 0; i < n1; ++ i) { cin >> tmp; if( ! i) { h1 = new ListNode( tmp); head1 = h1; } else { h1 -> next = new ListNode( tmp); h1 = h1 -> next; } } for( int i = 0; i < n2; ++ i) { cin >> tmp; if( ! i) { h2 = new ListNode( tmp); head2 = h2; } else { h2 -> next = new ListNode( tmp); h2 = h2 -> next; } } Solution solution; ListNode * ans = solution . addTwoNumbers( head1 , head2); puts( "----------------------"); while( ans) { cout << ans -> val; ans = ans -> next; } cout << endl; } return 0; }