刷LeetCode（2）——Add Two Numbers

Code it now! https://leetcode.com/problems/add-two-numbers/description/

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8

思路一：刚看到题目的时候，没考虑太多，最先想到的思路就是把链接转换为整数，然后求和，然后再把结果分配到新的链表中，结果实现如下：

#include <iostream>#include <string>#include <vector>using namespace std;/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        int num = 0;        ListNode *node = NULL;        long num1 = getNumberFromList(l1);        long num2 = getNumberFromList(l2);        long sum = num1 + num2;        if( !sum )        {            node = new ListNode(0);            return node;        }        while( sum )        {            num = sum % 10 ;            sum /= 10;            if( !node ){                node = new ListNode(num);            }            else            {                ListNode *p = new ListNode(num);                addNodeToTail(node,p);            }        }        return node;    }    long getNumberFromList(ListNode* l)    {        long order = 1;        long result = 0;         ListNode *node = NULL;        for( node = l; node ; node = node->next )        {            result += order * node->val;            order *= 10;        }        return result;    }    int addNodeToTail(ListNode *h,ListNode *node)    {        ListNode *last = NULL;        ListNode *cur = h;        if( !h || !node )        {            return -1;        }        while( cur )        {            last = cur;            cur = cur->next;        }        last->next = node;        node->next = NULL;        return 0;    }};int main(){    ListNode *node = NULL;    ListNode *result = NULL;    Solution    solution;    ListNode *l1 = new ListNode(9);    ListNode *l2 = new ListNode(1);    ListNode *node1 = new ListNode(9);    ListNode *node2 = new ListNode(9);    ListNode *node3 = new ListNode(9);    ListNode *node4 = new ListNode(9);    ListNode *node5 = new ListNode(9);    ListNode *node6 = new ListNode(9);    ListNode *node7 = new ListNode(9);    ListNode *node8 = new ListNode(9);    ListNode *node9 = new ListNode(9);    solution.addNodeToTail(l2,node1);    solution.addNodeToTail(l2,node2);    solution.addNodeToTail(l2,node3);    solution.addNodeToTail(l2,node4);    solution.addNodeToTail(l2,node5);    solution.addNodeToTail(l2,node6);    solution.addNodeToTail(l2,node7);    solution.addNodeToTail(l2,node8);    solution.addNodeToTail(l2,node9);    result = solution.addTwoNumbers(l1,l2);    for( node = result; node ; node = node->next )    {        cout << " " << node->val << " ";    }    cout << endl;    return 0;}

提交上去之后报如下图所示的错误：

看来是我想太简单了，如果数组过长的话，则会直接超过范围。

思路二：有上面的尝试，肯定不能简单的算数相加，采用轮询两个数组。具体实现如下：

#include <iostream>#include <string>#include <vector>using namespace std;/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        int carrybit = 0;        int sum =0;        ListNode *cur1 = NULL,*cur2 = NULL,*node = NULL;        for(cur1 = l1,cur2 = l2; cur1 && cur2 ; cur1 = cur1->next,cur2 = cur2->next)        {            sum = cur1->val + cur2->val + carrybit;            //cout << " cur1 = " << cur1->val << " cur2 = " << cur2->val << endl;            if( sum >= 10 )            {                sum -= 10;                carrybit = 1;            }            else            {                carrybit = 0;            }            addNumToList(&node,sum);        }        calculateTheRestList(node,cur1,&carrybit);        calculateTheRestList(node,cur2,&carrybit);        if( carrybit ){            addNumToList(&node,carrybit);            carrybit = 0;        }        return node;    }    void calculateTheRestList(ListNode *node,ListNode *cur,int *carrybit)    {        int sum = 0;        if( cur )        {            for( ListNode *q = cur; q ; q=q->next )            {                sum = q->val + *carrybit ;                if( sum >= 10 )                {                    sum -= 10;                    *carrybit = 1;                }                else                {                    *carrybit = 0;                }                addNumToList(&node,sum);            }        }    }    void addNumToList(ListNode **h,int num)    {        if( !(*h) ){            *h = new ListNode(num);        }        else        {            ListNode *p = new ListNode(num);            addNodeToTail(*h,p);        }    }    int addNodeToTail(ListNode *h,ListNode *node)    {        ListNode *last = NULL;        ListNode *cur = h;        if( !h || !node )        {            return -1;        }        while( cur )        {            last = cur;            cur = cur->next;        }        last->next = node;        node->next = NULL;        return 0;    }};int main(){    ListNode *node = NULL;    ListNode *result = NULL;    Solution    solution;    ListNode *l1 = new ListNode(9);    ListNode *l2 = new ListNode(1);    ListNode *node1 = new ListNode(9);    ListNode *node2 = new ListNode(9);    ListNode *node3 = new ListNode(9);    ListNode *node4 = new ListNode(9);    ListNode *node5 = new ListNode(9);    ListNode *node6 = new ListNode(9);    ListNode *node7 = new ListNode(9);    ListNode *node8 = new ListNode(9);    ListNode *node9 = new ListNode(9);    solution.addNodeToTail(l2,node1);    solution.addNodeToTail(l2,node2);    solution.addNodeToTail(l2,node3);    solution.addNodeToTail(l2,node4);    solution.addNodeToTail(l2,node5);    solution.addNodeToTail(l2,node6);    solution.addNodeToTail(l2,node7);    solution.addNodeToTail(l2,node8);    solution.addNodeToTail(l2,node9);    result = solution.addTwoNumbers(l1,l2);    for( node = result; node ; node = node->next )    {        cout << " " << node->val << " ";    }    cout << endl;    return 0;}

Leetcode统计的效率还不错：

优化：上面的实现比较直观，但是太多函数调用，效率也不高，于是自己优化一把，最终的代码如下：

#include <iostream>#include <string>#include <vector>using namespace std;/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        int carrybit = 0;        int sum =0;        ListNode *cur1 = l1,*cur2 = l2,*node = NULL,*last =NULL;        while( cur1 || cur2 )        {            sum = carrybit;            if( cur1 )            {                sum += cur1->val;                cur1 = cur1->next;            }            if( cur2 )            {                sum += cur2->val;                cur2 = cur2->next;            }            if( sum >= 10 )            {                sum -= 10;                carrybit = 1;            }            else            {                carrybit = 0;            }            if( !node )            {                node = new ListNode(sum);                last = node;            }            else            {                ListNode *p = new ListNode(sum);                last->next = p;                last = p;            }        }        if( carrybit ){            // the last point must no be NULL            ListNode *p = new ListNode(carrybit);            last->next = p;            last = p;            carrybit = 0;        }        return node;    }    int addNodeToTail(ListNode *h,ListNode *node)    {        ListNode *last = NULL;        ListNode *cur = h;        if( !h || !node )        {            return -1;        }        while( cur )        {            last = cur;            cur = cur->next;        }        last->next = node;        node->next = NULL;        return 0;    }};int main(){    ListNode *node = NULL;    ListNode *result = NULL;    Solution    solution;    ListNode *l1 = new ListNode(9);    ListNode *l2 = new ListNode(1);    ListNode *node1 = new ListNode(9);    ListNode *node2 = new ListNode(9);    ListNode *node3 = new ListNode(9);    ListNode *node4 = new ListNode(9);    ListNode *node5 = new ListNode(9);    ListNode *node6 = new ListNode(9);    ListNode *node7 = new ListNode(9);    ListNode *node8 = new ListNode(9);    ListNode *node9 = new ListNode(9);    solution.addNodeToTail(l2,node1);    solution.addNodeToTail(l2,node2);    solution.addNodeToTail(l2,node3);    solution.addNodeToTail(l2,node4);    solution.addNodeToTail(l2,node5);    solution.addNodeToTail(l2,node6);    solution.addNodeToTail(l2,node7);    solution.addNodeToTail(l2,node8);    solution.addNodeToTail(l2,node9);    result = solution.addTwoNumbers(l1,l2);    for( node = result; node ; node = node->next )    {        cout << " " << node->val << " ";    }    cout << endl;    return 0;}

结果网站上统计的效率还不如我优化之前，很是想不明白，各位如果有什么高见，一起讨论下。