兩個鏈表相加

1 題目
You are giventwo linked lists representing two non-negative numbers. The digits are storedin reverse order
and each of their nodes contain a single digit. Add the twonumbers and return it as a linked list.
Input: (2 -> 4 -> 3)+ (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
2 分析
該題目屬於鏈表的相加，須要注意的有：
(1) 考慮兩個鏈表的長度，尤爲是鏈表爲空時也能處理。
(2) 每一個結點只表示一位數字。
(3) 當鏈表末尾結點相加後如有進位，則須要申請新的結點存儲信息。

3 核心代碼：

int listLength(ListNode* head)  //遞歸獲得鏈表的長度
{  
    return head ? 1 + listLength(head ->next) : 0;  
}  
   
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2){  //把兩個鏈表相加(前面的參數爲長鏈表)
    if (listLength(l1) < listLength(l2)){   
        return addTwoNumbers(l2, l1);  
    }  
    ListNode *head1 = l1, *head2 = l2;  
    int inc = 0;  
    bool isEnd = false;  
    while (head2){ 
        int val = head1 -> val + head2-> val + inc;  
        head1 -> val = val % 10;  
        inc = val / 10;  
        if (head1 -> next){  
            head1 = head1 -> next;  
        }else{  
            isEnd = true;  
        }  
        head2 = head2 -> next;  
    }  
    while (inc){ //當短鏈表計算完以後
        int val = isEnd ? inc : head1 ->val + inc;  
        if (isEnd){   
            head1 -> next = new ListNode(val % 10);  
        }else{  
            head1 -> val = val % 10;  
        }  
        inc = val / 10;  
        if (head1 -> next){  
            head1 = head1 -> next;  
        }else{   
            isEnd = true;  
        }  
    }  
    return l1;  
}

View Code

4 如下爲代碼的完整實現

#include "stdafx.h"
#include<string.h>
#include<iostream>

using namespace std;
struct ListNode {  //sizeof(ListNode):8
    int val; 
    ListNode *next; 
    ListNode(int x) : val(x), next(NULL) {} 
}; 
 
   
int listLength(ListNode* head)  //遞歸獲得鏈表的長度
{  
    return head ? 1 + listLength(head ->next) : 0;  
}  
   
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2){  //把兩個鏈表相加(前面的參數爲長鏈表)
    if (listLength(l1) < listLength(l2)){   
        return addTwoNumbers(l2, l1);  
    }  
    ListNode *head1 = l1, *head2 = l2;  
    int inc = 0;  
    bool isEnd = false;  
    while (head2){ 
        int val = head1 -> val + head2-> val + inc;  
        head1 -> val = val % 10;  
        inc = val / 10;  
        if (head1 -> next){  
            head1 = head1 -> next;  
        }else{  
            isEnd = true;  
        }  
        head2 = head2 -> next;  
    }  
    while (inc){ //當短鏈表計算完以後
        int val = isEnd ? inc : head1 ->val + inc;  
        if (isEnd){   
            head1 -> next = new ListNode(val % 10);  
        }else{  
            head1 -> val = val % 10;  
        }  
        inc = val / 10;  
        if (head1 -> next){  
            head1 = head1 -> next;  
        }else{   
            isEnd = true;  
        }  
    }  
    return l1;  
}
//******************************************************************
ListNode *createList(int arr[],int len){  //根據數組建立鏈表
    ListNode* head = NULL,*p;
    p=head;
    for(int i=0; i<len; i++){
        if(i==0){
            head = new ListNode(arr[i]);
            p = head;
        }else{
            p->next = new ListNode(arr[i]);
            p = p->next;
        }
    }
    return head;
}
void printList(ListNode *head){ //輸出鏈表
    ListNode *p = head;
    if(head==NULL) cout<<"空鏈表"<<endl;
    else{
        while(p){
            cout<<p->val<<" ";
            p = p->next;
        }
        cout<<endl;
    }
}
int main(int argc, char * argv[])
{   //32位系統中全部指針(包括複雜結構體)均爲4個字節，float、long和int類型爲4個字節，char爲1個字節，double爲8個字節
    //char arr[] = "12345678";        //strlen:8  sizeof:9  sizeof(arr)/sizeof(arr[0]):9
                      //做爲參數時      strlen:8  sizeof:4  sizeof(arr)/sizeof(arr[0]):4
    //int arr[] = {1,2,3,4,5,6,7,8};  //strlen:x  sizeof:32  sizeof(arr)/sizeof(arr[0]):8
                      //做爲參數時      strlen:x  sizeof:4  sizeof(arr)/sizeof(arr[0]):1

    int arr1[] = {2, 4, 3};  //表示從低位到高位
    int arr2[] = {5, 6, 4};
    int len1 = sizeof(arr1)/sizeof(arr1[0]);
    int len2 = sizeof(arr2)/sizeof(arr2[0]);
    ListNode *l1 = createList(arr1,len1);
    ListNode *l2 = createList(arr2,len2);

    printList(addTwoNumbers(l1,l2));

    system("pause");
    return 0;
}

View Code