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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;
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原文地址：http://www.cnblogs.com/rangozhang/p/4550576.html
