前言：這是一道很有意思的題目，原題如下：

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

Input: (7 -> 1 -> 6), (5 -> 9 -> 2). That is to say you have to calculate 617+295 and return output:

Output: 2 -> 1 -> 9

Follow UP:

Suppose the digits are store in forward order, repeat the above problems

Input: (6 -> 1 -> 7), (2 -> 9 -> 5). That is to say you have to calculate 617+295 and return output:

Output: 9 -> 1 -> 2

譯文：

你有兩個(gè)由單鏈表表示的數(shù)。每個(gè)結(jié)點(diǎn)代表其中的一位數(shù)字。數(shù)字的存儲(chǔ)是逆序的， 也就是說(shuō)個(gè)位位于鏈表的表頭。寫一函數(shù)使這兩個(gè)數(shù)相加并返回結(jié)果，結(jié)果也由鏈表表示。

例子：(7-> 1 -> 6), (5 -> 9 -> 2)

輸入：2 -> 1 -> 9

進(jìn)階：

考慮鏈表是反向的，比如：

例子：(6-> 1 -> 7), (2 -> 9 -> 5)

輸入：9 -> 1 -> 2

考慮遞歸法與非遞歸法實(shí)現(xiàn)原題和進(jìn)階問(wèn)題（Follow Up）。

非遞歸方法：

我們必須保存每次結(jié)果的carryOn即進(jìn)位值，在第一種情況下，我們需要知道，表頭的數(shù)據(jù)是最低位，這意味著從表頭即可開(kāi)始加兩個(gè)數(shù)，并且把carryOn （1，或0）合理的傳遞給下一次運(yùn)算。

首先建立鏈表類如下：

private static class LinkedList{

private Node head;

public LinkedList (){

this.head = null;

}

public void insertNode (int value){

Node newNode = new Node(value);

newNode.prev = null;

newNode.next = this.head;

if (this.head!=null) this.head.prev = newNode;

this.head = newNode;

}

public void deleteNode(Node tobeDel){

//System.out.println("deleting:value["+tobeDel.value+"]"+tobeDel.next);

if (tobeDel == this.head) {

head = tobeDel.next;

head.prev = null;

}

if (tobeDel.prev!=null) tobeDel.prev.next = tobeDel.next;

if (tobeDel.next!=null) tobeDel.next.prev = tobeDel.prev;

}

public void printAllNodes(){

Node newNode = this.head;

while (newNode!=null){

if (newNode == head)

? ?System.out.print("[Head]"+newNode.value);

else

System.out.print("->"+newNode.value);

newNode = newNode.next;

}

System.out.println("[End]");

}

}

注意我同時(shí)建立了函數(shù)printAllNodes為了方便的打印所有鏈表。

下面是Node的類：

private static class Node{

private int value;

private Node next;

private Node prev;

public Node (int value){

this.value = value;

this.prev = null;

this.next = null;

}

}

核心代碼：

private static void CC2_5_1() {

// TODO Auto-generated method stub

LinkedList list1 = new LinkedList();

LinkedList list2 = new LinkedList();

LinkedList newList = new LinkedList();

list1.insertNode(6);

list1.insertNode(1);

list1.insertNode(7);

list2.insertNode(2);

list2.insertNode(9);

list2.insertNode(5);

list1.printAllNodes();

list2.printAllNodes();

Node point1 = list1.head;

Node point2 = list2.head;

int value=0,carryOn=0;

while (point1 !=null && point2 !=null){

value = point1.value + point2.value + carryOn;

carryOn = 0;

if (value >= 10){

value = value%10;

carryOn = 1;

}

newList.insertNode(value);

point1 = point1.next;

point2 = point2.next;

}

point1 = newList.head;

while (point1!=null){

newList.insertNode(point1.value);

newList.deleteNode(point1);

point1 = point1.next;

}

newList.printAllNodes();

}

遞歸方法實(shí)現(xiàn)FollowUp問(wèn)題

當(dāng)鏈表反轉(zhuǎn)，即表尾變成了加運(yùn)算的最低位時(shí)，上述方法變的相對(duì)麻煩很多，這時(shí)遞歸方法顯得更加方便。

核心代碼：

private static int addingTwoLinkedList(LinkedList newList,Node n1, Node n2){

int value = n1.value + n2.value;

if (n1.next ==null && n2.next == null){

newList.insertNode(value%10);

return value/10;

}

else{

value = value + addingTwoLinkedList(newList,n1.next,n2.next);

newList.insertNode(value%10);

return value/10;

}

}

函數(shù)addingTwoLinkedList的返回值是向其上層（previous Node）進(jìn)位的數(shù)值，非0即1. 函數(shù)的參數(shù)需要輸入兩個(gè)被加數(shù)的表頭即可。

遞

轉(zhuǎn)載于:https://blog.51cto.com/jamesd1987/1349625

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