2_兩數相加
描述
給定兩個非空鏈表來表示兩個非負整數。位數按照逆序方式存儲,它們的每個節點只存儲單個數字。將兩數相加返回一個新的鏈表。
你可以假設除了數字 0 之外,這兩個數字都不會以零開頭。
示例:
輸入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
輸出:7 -> 0 -> 8
原因:342 + 465 = 807
解法:小學數學
思路
這道題的求解思路和在紙上求兩數之和一樣,從最低位(鏈表表頭)開始,逐位(節點)求和,用變量 carry
保存前一位的進位結果,如果求和的過程到達長度較短的鏈表的尾部(空節點 null
),則求和時將該節點的值當作 0,求和結束後還需要判斷 carry
變量的值是否爲 1,如果爲 1 還需要添加一個節點(節點取值爲 1)作爲最高位。具體的求解過程如下圖所示:
非遞歸實現
Java 實現(非遞歸寫法 1)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(-1);
ListNode curNode = dummyHead, p = l1, q = l2;
int carry = 0;
while (p != null || q != null) {
int x = (p != null) ? p.val : 0;
int y = (q != null) ? q.val : 0;
int sum = x + y + carry;
carry = sum / 10;
curNode.next = new ListNode(sum % 10);
curNode = curNode.next;
if (p != null) {
p = p.next;
}
if (q != null) {
q = q.next;
}
}
if (carry > 0) {
curNode.next = new ListNode(carry);
}
return dummyHead.next;
}
}
複雜度分析:
- 時間複雜度:,其中 和 分別表示兩個鏈表的長度
- 空間複雜度:,返回鏈表的長度最多爲
Java 實現(非遞歸寫法 2)★
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
int carry = 0;
for (ListNode p1 = l1, p2 = l2; p1 != null || p2 != null; ) {
int sum = carry;
sum += (p1 != null) ? p1.val : 0;
sum += (p2 != null) ? p2.val : 0;
tail.next = new ListNode(sum % 10);
tail = tail.next;
carry = sum / 10;
p1 = (p1 == null) ? p1 : p1.next;
p2 = (p2 == null) ? p2 : p2.next;
}
if (carry != 0) {
tail.next = new ListNode(carry);
}
return dummy.next;
}
}
複雜度分析同上。
Python 實現
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
dummy_head = ListNode(-1)
cur = dummy_head
carry = 0
while l1 or l2 or carry:
v1 = v2 = 0
if l1:
v1 = l1.val
l1 = l1.next
if l2:
v2 = l2.val
l2 = l2.next
carry, val = divmod(v1 + v2 + carry, 10)
cur.next = ListNode(val)
cur = cur.next
return dummy_head.next
複雜度分析同上。
遞歸實現
Java 實現
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
return addTwoNumbers(l1, l2, 0);
}
private ListNode addTwoNumbers(ListNode l1, ListNode l2, int carry) {
// Recursive termination condition
if (l1 == null && l2 == null) {
return carry > 0 ? new ListNode(carry) : null;
}
int sum = carry;
ListNode l1Next = null, l2Next = null;
if (l1 != null) {
sum += l1.val;
l1Next = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2Next = l2.next;
}
ListNode curr = new ListNode(sum % 10);
curr.next = addTwoNumbers(l1Next, l2Next, sum / 10);
return curr;
}
}
複雜度分析:
- 時間複雜度:,其中 和 分別表示兩個鏈表的長度
- 空間複雜度:,額外空間是由於遞歸調用佔用系統棧的空間,遞歸的深度最多爲 層
Python 實現
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
def helper(l1, l2, carry):
if not l1 and not l2:
if carry > 0:
return ListNode(carry)
else:
return None
sum = carry
l1_next, l2_next = None, None
if l1:
sum += l1.val
l1_next = l1.next
if l2:
sum += l2.val
l2_next = l2.next
curr = ListNode(sum % 10)
curr.next = helper(l1_next, l2_next, sum // 10)
return curr
return helper(l1, l2, 0)
複雜度分析同上。