leetcode 中等题

2. Add Two Numbers（中等）

/**
 * 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) {
        ListNode* first = l1;
        ListNode* second = l2;
        int result1=0,result2=0;
        long long result=0;
        stack<int> S;
        int count=0;
        
        while(l1 != NULL){
            S.push(l1->val);
            l1 = l1->next;
        }
            
            
        count = S.size()-1;
        while(!S.empty()){
            result1 += pow(10,count--) * S.top();
            S.pop();
        }
            
        
        while(l2 != NULL){
            S.push(l2->val);
            l2 = l2->next;
        }
            
        count = S.size()-1;
        while(!S.empty()){
            result2 += pow(10,count--) * S.top();
            S.pop();
        }
            
        result = result1 + result2;
        delete first,second;
        
        // ListNode*p = new ListNode(-1);  //头节点
        ListNode* pHead = NULL;
        ListNode* p = pHead;
        
        if(!result){
            ListNode* q = new ListNode(0);
            pHead = q;
        }
        while(result){
            int val = result % 10;
            result /= 10;
            
            ListNode* q = new ListNode(val);
            if(pHead == NULL){
                pHead = q;
                p = q;
            }
            else{
                p->next = q;
                p = p->next;
            }
        }
        return pHead;
    }
};

此方法会溢出，超过long long 型范围

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {   //正确解法，按位相加，注意进位
    ListNode *head = NULL, *prev = NULL;
    int carry = 0;
    while (l1 || l2) {
        int v1 = l1? l1->val: 0;
        int v2 = l2? l2->val: 0;
        int tmp = v1 + v2 + carry;
        carry = tmp / 10;
        int val = tmp % 10;
        ListNode* cur = new ListNode(val);
        if (!head) head = cur;
        if (prev) prev->next = cur;
        prev = cur;
        l1 = l1? l1->next: NULL;
        l2 = l2? l2->next: NULL;
    }
    if (carry > 0) {
        ListNode* l = new ListNode(carry);
        prev->next = l;
    }
    return head;
}

 3， Longest Substring Without Repeating Characters.（很慢）

int lengthOfLongestSubstring(string s) {

    int maxLength = 0;
    int count;
    for(int i=0;i<s.length();i++){
        string currString = "";
        count = 0;
        for(int j=i;j<s.length();j++){
            if(currString.find(s[j])>=currString.length()){
                currString += s[j];
                count++;
                cout<<s[j]<<endl;
                cout<<"string:"<<currString<<endl; 
                if(count>maxLength) maxLength = count;
            } 
            else{
                break;
            }
        }
    }
    return maxLength;
}

5， Longest Palindromic Substring.（很慢）

string longestPalindrome(string s) {
    string maxString = "";
    int maxLength = 0;
    int count = 0;
        
    for(int i=0;i<s.length();i++){
        string Str = "";
        int index = s.rfind(s[i]);
        int start = i;
        int end = index;
        int count = 0;
        int ii=start;
            
        while(ii<=index){
            if(s[ii]==s[index]){ii++;index--;}
            else{
                ii = start;
                index = end-1;
                end--;
            }   //匹配失败
        }
            
        if(ii>index){  //匹配成功
            for(int k=start;k<=end;k++)
                Str += s[k];
            count = Str.length();
        }
        if(count>=maxLength) {maxString = Str;maxLength=count;}
    }
    return maxString;
}