Leetcode(5)-Add two numbers

2 Add two numbers
模拟手工加和的过程即可，结构体中提供了构造函数；需要注意的是两个链表可能不等长，需要分别处理剩下的位数，还有最高位加和可能存在进位也需处理。

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if (l1 == NULL) return l2;  
        if (l2 == NULL) return l1;
        int c=0;//进位
        ListNode* ret=NULL,*pNext=NULL,*pNode=NULL;
        while(l1!=NULL && l2!=NULL){
            //使用构造函数构造新节点
            pNext = new ListNode(l1->val + l2->val + c);  
            c = pNext->val / 10;    //计算进位 
            pNext->val = pNext->val % 10;   //计算该位的数字 
            if (ret == NULL)  //首个节点 
                ret = pNode = pNext;    
            else //头结点不为空 
            {  
                pNode->next = pNext;  
                pNode = pNext;  
            }  
            l1 = l1->next;  
            l2 = l2->next;  
        }
        //处理链表l1剩余的高位 
        while (l1 != NULL)  
        {  
            pNext = new ListNode(l1->val + c);  
            c = pNext->val / 10;      
            pNext->val = pNext->val % 10;  
            pNode->next = pNext;  
            pNode = pNext;  
            l1 = l1->next;  
        }  
        //处理链表l2剩余的高位 
        while (l2 != NULL)  
        {  
            pNext = new ListNode(l2->val + c);  
            c = pNext->val / 10;      
            pNext->val = pNext->val % 10;  
            pNode->next = pNext;  
            pNode = pNext;  
            l2 = l2->next;  
        }  
        //如果有最高处的进位，需要增加结点存储 
        if (c > 0)  
        {  
            pNext = new ListNode(c);  
            pNode->next = pNext;  
        }  
        return ret;
    }
};

3.Longest Substring Without Repeating Characters

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int n = s.length();
        int i = 0, j = 0;
        int maxLen = 0;
        //字母ascii码94
        bool exist[256] = {false};
        while (j < n) {
            //当遇到重复字母时，可能存在比已知更长的串
            if (exist[s[j]]) {
                maxLen = max(maxLen, j-i);
                while (s[i] != s[j]) {
                    exist[s[i]] = false;
                    i++;
                }
            i++;
            j++;
            //否则标记
            } else {
                exist[s[j]] = true;
                j++;
            }
        }
        //别忘了处理剩余字串
        maxLen = max(maxLen, n-i);
        return maxLen;
    }
};