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LeetCode 445_Add 2 Numbers_2.ts
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LeetCode 445_Add 2 Numbers_2.ts
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/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
/**
* Leet Code 445.
*
* Solution uses an O(1) space aimed to reduce the required space for reversal or virtual reversal. It rather uses
* the length of the linked list to identify when a padding can be used and when can it be skipped. This step is done
* without the help of additional carry, since carry needs to be added in the reverse order.
*
* To adjust the carry pointer, we need to reverse the list, adjust the carry and reverse again.
*
* O(n) time for run-time, O(1) for space.
*
* @param l1
* @param l2
*/
function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
// Finding the length of each linked list.
let l1Length = length(l1);
let l2Length = length(l2);
let resultCurr: ListNode | null = new ListNode(0, null), resultHead = resultCurr, lastCurr: ListNode | null = resultCurr;
// Adding items with prefixes attached, not using the carry.
while(l1Length >= 0 && l2Length >= 0 && l1 !== null && l2 !== null) {
if (l1Length > l2Length) {
// L1 has more items than l2, lets pull from l1 and add 0 for l2.
resultCurr.val = l1.val;
l1 = l1.next;
resultCurr.next = new ListNode(0, null)
l1Length--;
} else if (l1Length < l2Length) {
// L2 has more items than l1, lets pull from l2 and add 0 for l1.
resultCurr.val = l2.val;
l2 = l2.next;
resultCurr.next = new ListNode(0, null)
l2Length--
} else {
// Both have equal items, lets add.
resultCurr.val = l2.val + l1.val;
l2 = l2.next; l1 = l1.next;
resultCurr.next = new ListNode(0, null)
l2Length--; l1Length--;
}
lastCurr = resultCurr;
resultCurr = resultCurr.next;
}
lastCurr.next = null;
let newHead = reverseLinkedList(resultHead);
// Adjust the carry now.
adjustCarry(newHead)
// Reverse the list and return.
newHead = reverseLinkedList(newHead);
return newHead;
};
function adjustCarry(list: ListNode | null) {
let curr = list, last = null, carry = 0;
while(curr !== null) {
const val = curr.val + carry;
curr.val = val % 10;
carry = Math.floor(val / 10)
last = curr;
curr = curr.next;
}
// Add the pending carry to a new node.
if (carry > 0 && last !== null) {
last.next = new ListNode(carry, null);
}
}
function length(l1: ListNode | null) {
let length = 0, curr = l1;
while(curr !== null) {
length++
curr = curr.next;
}
return length;
}
function reverseLinkedList(l1: ListNode | null) {
let curr = l1, last: ListNode | null = null;
while(curr !== null) {
const next = curr.next;
curr.next = last;
last = curr;
curr = next;
}
return last;
}