本文共 2998 字，大约阅读时间需要 9 分钟。

老姐让我刷LeetCode的题，就当复习数据结构了，算法和代码能力真的只有通过刷题才提升得快啊

通过第一次的题也大概知道了leetcode的写题方式，不用头文件，库文件各种东西它只给你一个函数/方法, 把这个题的解决方法写入这个函数/方法即可，而且支持vim模式下的编辑，真的很方便

*

Given an array of integers, return indices of the two numbers such that theyadd up to a specific target.You may assume that each input would have exactly one solution, and you maynot use the same element twice.

• Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

Mycode:

class Solution {public:    vector
   
     twoSum(vector
    
     & nums, int target) {        vector
     
       ans;        int l=nums.size();        bool flag=false;        for(int i=0;i
     
    
   

Given a 32-bit signed integer, reverse digits of an integer.

• Example 1:

Input: 123Output: 321

• Example 2:

Input: -123Output: -321

• Example 3:

Input: 120Output: 21

• Note:

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31,  2^31 − 1]. For the purposeof this problem, assume that your function returns 0 when the reversedinteger overflows.

Mycode:

class Solution {public:    int reverse(int x) {	    long long ans=0;        long long Max = ((long long)1<<31) - 1;        long long Min = -(Max+1);        if (x==0){            ;        }        else{            vector
   
     v;	        while(x){		        v.push_back(x%10);		        x /= 10;	        }	        for(int i = 0; i < v.size()-1; i++){		        ans = (ans + v[i])*10;	        }	        ans += v.back();        }        if (ans > Max || ans < Min) ans=0;	    return (int)ans;    }};
   

这道题简单，但是坑点贼多，我WA了好多发，但主要还是自己写程序bug太多了啊。

记录一下心得吧。1. int范围能撑到1<<30，1<<31要用long long定义.2. 初始化1<<31这种long long型数据时，要把1强转为long long才行，要不然会溢出. 3. 预算符优先级：+,- > <<,>>

You are given two non-empty linked lists representing two non-negativeintegers. The digits are stored in reverse order and each of their nodescontain a single digit. Add the two numbers and return it as a linked list.You may assume the two numbers do not contain any leading zero, except thenumber 0 itself.

• Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.

代码已经注释的很详细了!

/** * 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 *ans = new ListNode(-1);   //创建答案结点        ListNode *p = ans;        int carry=0;    //记录进位        while(l1 || l2){    //l1,l2链表全部遍历完才退出循环            int x = l1 ? l1->val : 0;            int y = l2 ? l2->val : 0;            int sum = x + y + carry;            carry = sum/10;     //更新进位,若有进位，则给下一个l1,l2结点加起来的sum+1            p->next = new ListNode(sum % 10);//更新p的下一个结点            p = p->next;            if (l1) l1 = l1->next;            if (l2) l2 = l2->next;        }        //特殊情况处理，若l1,l2等长且最后一个结点相加>9，则再进一次位        if (carry) p->next = new ListNode(1);        return ans->next;    }};

转载地址：http://defqb.baihongyu.com/