LeetCode (Two Sum & Add Two nums)

还是觉得有必要坚持刷leetcode   在这里打卡。 加油。    这是第一遍  很惭愧。做这一行这么久 才开始刷力扣。但是开始了总是好的。尽管代码丑,但是至少都是能通过的吧。 

 

1、 Total Accepted: 159385 Total Submissions: 822358 Difficulty: Medium

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

My Answer：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        
        int temp,temp_j;
        map<int,int> num_map;
        vector<int> ret;
        for(int i=0;i<nums.size();++i){
            
            temp = nums.at(i);
            temp_j = target-temp;
            if( num_map.count(temp_j)){
            
                if(i > num_map[temp_j]){
                    ret.push_back(num_map[temp_j]+1);
                    ret.push_back(i+1);
                    return ret;
                }
            }
            num_map.insert( pair<int,int>(temp,i) );
        }
    }
};

 

2、You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

My Answer:  

/**
 * 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 *t_l1 = l1;
        ListNode *t_l2 = l2;
        ListNode *p=NULL;
        ListNode *l = t_l1;
        
        bool inc_ = false;
        while( t_l1!=NULL || t_l2!=NULL ){
            if(t_l1){
                p = t_l1;
            }else{            
                if(p){
                    p->next = t_l2;
                }
                p = t_l2;
            }
                
            if(t_l1&&t_l2){
                t_l1->val = t_l1->val+t_l2->val;                
                if( t_l1->val>=10 ){
                    inc_ = true;
                    t_l1->val = t_l1->val%10;          
                }
                t_l1 = t_l1->next;
                t_l2 = t_l2->next;
            }else if(t_l1){                
                if( t_l1->val>=10 ){
                    inc_ = true;
                    t_l1->val = t_l1->val%10;
                }
                t_l1 = t_l1->next;
            }else if(t_l2){                
                t_l1 = t_l2;                
                if( t_l1->val>=10 ){
                    inc_ = true;
                    t_l1->val = t_l1->val%10;
                }
                t_l1 = t_l1->next;
                t_l2 = NULL;
            }
            
            if( inc_ ){
                if(!t_l1){
                    t_l1 = new ListNode(0);
                    p->next = t_l1;
                }
                
                t_l1->val = t_l1->val+1;
                inc_ = false;
            }
            
        }
     　　return l;
    }
};