LeetCode 2. add two numbers

看似简单，轻易AD，但重点是和其他人写的代码进行比较从而学习

C++:

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

        ListNode* result;

        ListNode* cur;

        ListNode* cur_1 = l1->next, *cur_2 = l2->next;

        

        int cur_value = l1->val + l2->val;

        int hasOne = cur_value/10;

//      ListNode first();  因为用了cur=&first而不是new导致Runtime Error,个人认为具体原因是出了函数first就被当做临时变量释放

        cur = result = new ListNode(cur_value - hasOne*10);

        

        while (cur_1 != NULL && cur_2 != NULL) {

            cur_value = cur_1->val + cur_2->val + hasOne;

            hasOne = cur_value/10;

            cur->next = new ListNode(cur_value - hasOne*10);

            cur = cur->next;

            cur_1 = cur_1->next;

            cur_2 = cur_2->next;

        }

        

        while(cur_1 != NULL) {

            int tmp = cur_1->val;

            tmp += hasOne;

            hasOne = tmp/10;

            cur_1 = cur_1->next;

            cur->next = new ListNode(tmp-hasOne*10);

            cur = cur->next;

        }

        

        while(cur_2 != NULL){

            int tmp = cur_2->val;

            tmp += hasOne;

            hasOne = tmp/10;

            cur_2 = cur_2->next;

            cur->next = new ListNode(tmp-hasOne*10);

            cur = cur->next;

        }

        

        if (hasOne > 0) {

            cur->next = new ListNode(hasOne);

        }

        

        return result;

    }

笨到家的方法，显然是一个一个特殊情况想起来之后加进去，没有章法。通过别人代码改进后：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

        int hasOne = 0;

        ListNode preheader(-1);

        ListNode* cur = &preheader;

        while (l1 || l2 || hasOne) {

            int cur_value = (l1?l1->val:0) + (l2?l2->val:0) + hasOne;

            l1 = l1?l1->next:NULL;

            l2 = l2?l2->next:NULL;

            hasOne = cur_value/10;

            cur->next = new ListNode(cur_value%10);

            cur = cur->next;

        }

        return preheader.next;

    }

1. preheader解决了头结点特殊情况问题

2. 利用 1220 + 12 = 1220 + 0012的办法将多个情况分支融合

Python(自己写的形如c++，但对比方知python代码之简洁):

#from https://leetcode.com/discuss/36908/python-for-the-win

def addTwoNumbers(self, l1, l2): 
　　addends = l1, l2
　　dummy = end = ListNode(0)
　　carry = 0
　　while addends or carry:
　　　　carry += sum(a.val for a in addends)
　　　　addends = [a.next for a in addends if a.next]
　　　　end.next = end = ListNode(carry % 10)
　　carry /= 10
　　return dummy.next

#version of transfer to int then transfer back

def addTwoNumbers(self, l1, l2): 

    def toint(node): 

        return node.val + 10 * toint(node.next) if node else 0 

    def tolist(n): 

        node = ListNode(n % 10)

        if n > 9: 

            node.next = tolist(n / 10) 

        return node 

return tolist(toint(l1) + toint(l2))                

1. m = n = 1的情况下，n改变m不会跟着改变；m = n = My_Class的情况下，n改变m也会跟着变

2. end.next = end = Class()等价于end.next = Class(); end = end.next;