/*
* problem statement:
*
* You are given two linked lists representing two non-negative numbers.
* 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
*
*/
/*
* source code from blog:
* http://fisherlei.blogspot.ca/2013/01/leetcode-add-two-numbers-solution.html
*
* http://blog.csdn.net/lilong_dream/article/details/19544995
*
* analysis from the above blog:
* 要考虑进位。表示结果的结点要new出来。
*
* Julia's comment:
* 1. Write down the thought process
* First, step 1, base case, if one of list is empty, then, return the other one;
* step 2: do some calculation, get length of two list;
* step 3: define two explainable variable, one is called short one, one is longer one by comparing list length
* The code will be more clear to read, and discussion of addition is more clear to read.
* Adopt this smart idea!
* step 4: go through two lists in the same time, until short one finishes.
* first, calculate addition of two values (maybe 2 digits), and then, carry and value
* and then, store the sum in the new list; take care new list head creation, build up link
* step 5: go through the longer list alone,
* similar work as step 4
* step 6: if there is extra node to create for the sum, last carry.
*
* online judge:
* 1555 / 1555 test cases passed.
Status: Accepted
Runtime: 208 ms
*/
public static ListNode addTwoNumbers(ListNode l1, ListNode l2)
{
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if (l1 == null)
{
return l2;
}
if (l2 == null)
{
return l1;
}
int len1 = getLength(l1); // length of list 1
int len2 = getLength(l2);
ListNode longer = len1 >= len2 ? l1 : l2; // smart! reduce if discussion
ListNode shorter = len1 < len2 ? l1 : l2;
ListNode result = null;
ListNode sum = null;
int val = 0;
int carry = 0;
while (shorter != null)
{
val = longer.val + shorter.val + carry;
carry = getCarry(val); // at most two digit, get the second digit expressing 10; if there is one, it is 1
val = getValue(val); // adjust val to one digit
if (sum == null) // beginning of addition
{
sum = new ListNode(val);
result = sum;
}
else
{
sum.next = new ListNode(val);
sum = sum.next;
}
longer = longer.next;
shorter = shorter.next;
}
while (longer != null)
{
val = longer.val + carry;
carry = val / 10;
val -= carry * 10;
sum.next = new ListNode(val);
sum = sum.next;
longer = longer.next;
}
if (carry != 0)
{
sum.next = new ListNode(carry);
}
return result;
}
/*
* problem statement:
*
* You are given two linked lists representing two non-negative numbers.
* 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.
* Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
* Output: 7 -> 0 -> 8
*
*/
/*
* source code from blog:
* http://fisherlei.blogspot.ca/2013/01/leetcode-add-two-numbers-solution.html
*
* http://blog.csdn.net/lilong_dream/article/details/19544995
*
* analysis from the above blog:
* 要考虑进位。表示结果的结点要new出来。
*
* Julia's comment:
* 1. Write down the thought process
* First, step 1, base case, if one of list is empty, then, return the other one;
* step 2: do some calculation, get length of two list;
* step 3: define two explainable variable, one is called short one, one is longer one by comparing list length
* The code will be more clear to read, and discussion of addition is more clear to read.
* Adopt this smart idea!
* step 4: go through two lists in the same time, until short one finishes.
* first, calculate addition of two values (maybe 2 digits), and then, carry and value
* and then, store the sum in the new list; take care new list head creation, build up link
* step 5: go through the longer list alone,
* similar work as step 4
* step 6: if there is extra node to create for the sum, last carry.
*
* online judge:
* 1555 / 1555 test cases passed.
* Status: Accepted
* Runtime: 208 ms
*/
public static ListNode addTwoNumbers(ListNode l1, ListNode l2)
{
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if (l1 == null)
{
return(l2);
}
if (l2 == null)
{
return(l1);
}
int len1 = getLength(l1); // length of list 1
int len2 = getLength(l2);
ListNode longer = len1 >= len2 ? l1 : l2; // smart! reduce if discussion
ListNode shorter = len1 < len2 ? l1 : l2;
ListNode result = null;
ListNode sum = null;
int val = 0;
int carry = 0;
while (shorter != null)
{
val = longer.val + shorter.val + carry;
carry = getCarry(val); // at most two digit, get the second digit expressing 10; if there is one, it is 1
val = getValue(val); // adjust val to one digit
if (sum == null) // beginning of addition
{
sum = new ListNode(val);
result = sum;
}
else
{
sum.next = new ListNode(val);
sum = sum.next;
}
longer = longer.next;
shorter = shorter.next;
}
while (longer != null)
{
val = longer.val + carry;
carry = val / 10;
val -= carry * 10;
sum.next = new ListNode(val);
sum = sum.next;
longer = longer.next;
}
if (carry != 0)
{
sum.next = new ListNode(carry);
}
return(result);
}
public static int getLength(ListNode l)
{
if (l == null) return 0;
int len = 0;
ListNode head = l;
while (head != null) // get length of list 1
{
++len;
head = head.next;
}
return len;
}
