LeetCode OJ - Add Two Numbers (C++) (try 8 times)

為了練習尋找實習時會碰到的問題，我用別人推薦的 LeetCode online judge。
希望能直接寫程式，不用事先編譯就能直接AC。

To practice future interview question, I use LeetCode online judge.
I am trying to type code directly on website without compile in other tool.

編譯錯誤次數 Compile Error number: 3
嘗試次數 Try number:  8 (AC: 1) (WA:2) (CE:3) (RE: 2)
是否事先在其他工具編譯 if compile first in other tool: No

使用的程式語言 using programming language: C++
以前是否看過 seen this problem before: No

點這裡看題目 Click here to see this Problem!

1. use new(int val) to create new node and allocate space..
malloc is hard to use..

2. 在做 new的時候，要用nextNode -> next = new ListNode(tmpInt);
要是之前先去next, 就會變成NULL, 就會找不到指標了
ex:
nextNode = nextNode -> next;
nextNode = new ListNode(tmpInt);

	/**
	* 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) {
	ListNode *root = NULL;
	ListNode *nextNode = NULL;
	int carry = 0, tmpInt, num1, num2;
	//maybe l1 = NULL and l2 = NULL and carry = 1
	while(l1 != NULL || l2 != NULL || carry != 0){
	num1 = (l1 == NULL)? 0: l1 -> val;
	num2 = (l2 == NULL)? 0: l2 -> val;
	tmpInt = num1 + num2 + carry;
	carry = tmpInt/10; //(tmpInt >= 10)? 1: 0;
	tmpInt = tmpInt % 10;
	l1 = (l1 == NULL)? NULL: l1 -> next;
	l2 = (l2 == NULL)? NULL: l2 -> next;
	if(root == NULL){
	root = new ListNode(tmpInt);
	nextNode = root;
	}
	else{
	//if just use nextNode = new ListNode(tmpInt);
	//the previous nextNode is NULL, we will lost pointer
	nextNode -> next = new ListNode(tmpInt);
	nextNode = nextNode -> next;
	}
	}
	return root;
	}
	};

view raw LeetCode OJ - Add Two Numbers (try 8 times).cpp hosted with ❤ by GitHub

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