2. Two numbers added

2. Two numbers added

A two non-empty list is given to represent two non-negative integers. The digits are stored in reverse order, with each node storing only a single number. Adds two numbers to return a new linked list.

you can assume that except for the number 0, none of the two numbers will start with 0.

Example:

Input: (2, 4, 3) + (5, 6, 4) output: 7, 0,8 Cause:342 + 465 = 807

Solution Solutions

Method: Elementary Mathematics

Ideas

We use variables to keep track of carry and to simulate the process of adding bits from a header that contains the least significant bit.

Figure 1, visualization of the two-number addition method: 342 + 465 = 807342+465=807, each node contains a number, and the digits are stored in reverse order.

Algorithm

just as you calculate two numbers on a paper and that, we start by adding the least significant bit, which is the table header of the list l1l1 and l2l2. Since each digit should be at 0 \ldots 90 ... In the range of 9, we calculate the number of two and the "overflow" may occur. For example, 5 + 7 = 125+7=12. In this case, we will set the value of the current bit to 22 and carry carry = 1carry=1 into the next iteration. The carry Carrycarry must be 00 or 11, because two numbers are added (taking into account the rounding) that can occur for the largest and 9 + 9 + 1 = 199+9+1=19.

Please pay special attention to the following situations:

Test Cases	Description
l1=[0,1] l2=[0,1,2]	When a list is longer than another list.
L1=[] l2=[0,1]	When a list is empty, an empty list appears.
L1=[9,9] L2=[1]	The sum operation may end up with additional rounding, which can easily be forgotten

1 classSolution {2  Public:3listnode* addtwonumbers (listnode* L1, listnode*L2) {4ListNode *p1 = L1, *p2 =L2;5         intsum, CF, remain;6 7         //Digit8sum = P1->val + p2->Val;9CF = SUM/Ten;Tenremain = sum%Ten; One  AListNode *ret =NewListNode (remain); -ListNode *P3 =ret; -  theP1 = p1->Next; -P2 = p2->Next; -  -          while(P1! = NULL && P2! =NULL) +         { -sum = p1->val + P2->val +CF; +CF = SUM/Ten; Aremain = sum%Ten; at              -P3->next =NewListNode (remain); -P3 = p3->Next; -  -P1 = p1->Next; -P2 = p2->Next; in         } -  to          while(P1! =NULL) +         { -             if(cf! =0) the             { *sum = P1->val +CF; $CF = SUM/Ten;Panax Notoginsengremain = sum%Ten; -  theP3->next =NewListNode (remain); +P3 = p3->Next; AP1 = p1->Next; the             } +             Else -             { $P3->next =NewListNode (p1->val); $P3 = p3->Next; -P1 = p1->Next; -             } the         } -          while(P2! =NULL)Wuyi         { the             if(cf! =0) -             { Wusum = P2->val +CF; -CF = SUM/Ten; Aboutremain = sum%Ten; $  -P3->next =NewListNode (remain); -P3 = p3->Next; -P2 = p2->Next; A             } +             Else the             { -P3->next =NewListNode (p2->val); $P3 = p3->Next; theP2 = p2->Next; the             } the         } the  -         if(cf!=0) inP3->next =NewListNode (1); the  the         returnret; About     } the};

Analysis of Complexity:

L time complexity:O (\max (m, n)) O (max (m,n)), assuming mm and nn represent the lengths of L1l1 and L2L2 respectively, the algorithm above repeats at most \max (M, N) Max (m,n) times.

L Spatial complexity:O (\max (m, n)) O (max (m,n)), the new list has a maximum length of \max (m,n) + 1max (m,n) +1.

2. Two numbers added