# Two Lists Sum
### Source
- CC150 - [(167) Two Lists Sum](http://www.lintcode.com/en/problem/two-lists-sum/)
~~~
You have two numbers represented by a linked list, where each node contains a single digit.
The digits are stored in reverse order, such that the 1’s digit is at the head of the list.
Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given two lists, 3->1->5->null and 5->9->2->null, return 8->0->8->null
~~~
### 题解
一道看似简单的进位加法题，实则杀机重重，不信你不看答案自己先做做看。
首先由十进制加法可知应该注意进位的处理，但是这道题仅注意到这点就够了吗？还不够！因为两个链表长度有可能不等长！因此这道题的亮点在于边界和异常条件的处理，来瞅瞅我自认为相对优雅的实现。
### C++ - Iteration
~~~
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2) {
if (NULL == l1 && NULL == l2) {
return NULL;
}
ListNode *sumlist = new ListNode(0);
ListNode *templist = sumlist;
int carry = 0;
while ((NULL != l1) || (NULL != l2) || (0 != carry)) {
// padding for NULL
int l1_val = (NULL == l1) ? 0 : l1->val;
int l2_val = (NULL == l2) ? 0 : l2->val;
templist->val = (carry + l1_val + l2_val) % 10;
carry = (carry + l1_val + l2_val) / 10;
if (NULL != l1) l1 = l1->next;
if (NULL != l2) l2 = l2->next;
// return sumlist before generating new ListNode
if ((NULL == l1) && (NULL == l2) && (0 == carry)) {
return sumlist;
}
templist->next = new ListNode(0);
templist = templist->next;
}
return sumlist;
}
};
~~~
### 源码分析
1. 迭代能正常进行的条件为`(NULL != l1) || (NULL != l2) || (0 != carry)`, 缺一不可。
1. 对于空指针节点的处理可以用相对优雅的方式处理 - `int l1_val = (NULL == l1) ? 0 : l1->val;`
1. 生成新节点时需要先判断迭代终止条件 - `(NULL == l1) && (NULL == l2) && (0 == carry)`, 避免多生成一位数0。
### 复杂度分析
没啥好分析的，时间和空间复杂度均为 O(max(L1,L2))O(max(L1, L2))O(max(L1,L2)).
### C++ - Recursion
除了使用迭代，对于链表类问题也比较适合使用递归实现。
To-be done.
### Reference
- *CC150 Chapter 9.2* 题2.5，中文版 p123
- [Add two numbers represented by linked lists | Set 1 - GeeksforGeeks](http://www.geeksforgeeks.org/add-two-numbers-represented-by-linked-lists/)
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume