挑戰編程題（三）

接 挑戰編程題（二） http://www.cnblogs.com/qiange/p/5090588.html

 

題目：

2. 有兩個非負數的鏈表，每一個鏈表都反序的存儲一個多位數字的每一位，（例如：鏈表2->4->3  數字342），如今求兩個鏈表數字的相加，求和的鏈表。

      例如： Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

                  Output: 7 -> 0 -> 8

 

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

       //函數

    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

 

    }

 

 

解答：這道題不難主要是考察「單鏈表「的相關操做。

 解題思想，就是單鏈表每一個位置相加，若是每位和大約等於10，向前一位進一，依次類推；

以下：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

                ListNode *result=NULL;

                ListNode *l=result;

                 int carry =0;   //進位值

 

                 while (l1 && l2){

                                 int val=0;

                                 int tempVal= l1->val + l2->val + carry;

                                 if (tempVal>=10){

                                                val=tempVal-10;

                                                carry=1;

                                }

                                 else {

                                                val=tempVal;

                                                carry=0;

                                }

 

                                ListNode *l3=(ListNode *)malloc(

                                                 sizeof (ListNode));

                                l3->val=val;

                                l3->next = NULL;

                                 if (!l){

                                                l=l3;

                                                result=l;

                                } else {

                                                l->next=l3;

                                                l=l->next;

                                }

 

                                l1=l1->next;

                                l2=l2->next;

                }

                 //對l1鏈表剩餘項處理

                 while (l1){

                                 int val=0;

                                 int tempVal= l1->val + carry;

                                 if (tempVal>=10){

                                                val=tempVal-10;

                                                carry=1;

                                }

                                 else {

                                                val=tempVal;

                                                carry=0;

                                }

 

                                ListNode *l3=(ListNode *)malloc(

                                                 sizeof (ListNode));

                                l3->val=val;

                                l3->next = NULL;

                                 if (!l){

                                                l=l3;

                                } else {

                                                l->next=l3;

                                                l=l->next;

                                }

 

                                l1=l1->next;

                                

                }

 

                 //對l2鏈表剩餘項處理

                 while (l2){

                                 int val=0;

                                 int tempVal= l2->val + carry;

                                 if (tempVal>=10){

                                                val=tempVal-10;

                                                carry=1;

                                }

                                 else {

                                                val=tempVal;

                                                carry=0;

                                }

 

                                ListNode *l3=(ListNode *)malloc(

                                                 sizeof (ListNode));

                                l3->val=val;

                                l3->next = NULL;

                                 if (!l){

                                                l=l3;

                                } else {

                                                l->next=l3;

                                                l=l->next;

                                }

                                l2=l2->next;

                }

 

                 //對還存在的進位處理

                 if (carry)

                {

                                ListNode *l3=(ListNode *)malloc(

                                                 sizeof (ListNode));

                                l3->val=carry;

                                l3->next = NULL;

 

                                l->next=l3;

                                l=l->next;

                }

                 return result;

 

}

 

運行結果以下圖所示：（完整源碼下載地址： http://download.csdn.net/detail/cq20151207/9389996 ）

 

好久沒有用C++的單鏈表了，平時用php數組實現的不少東西，此次經過這道題，讓我有從新回顧了一下單鏈表的使用；

關於單鏈表的使用，我整理了一下，你們能夠參考下： http://www.cnblogs.com/qiange/p/5100441.html

 

今天的題目

 

3.給定一個字符串，得到最長且不重複的字符的子串的長度。例如，「abcabcbb」 的最長子串而沒必要重複字母是「abc」，它的長度是3爲「bbbbb」最長子串而沒必要重複字母是 「b」，它的長度是1。

 

 int lengthOfLongestSubstring(string s) {

 

    }

 

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