leetcode-day2

Reverse Intege

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

Example 1:
1
2
Input: 123
Output: 321
Example 2:
1
2
Input: -123
Output: -321
Example 3:
1
2
Input: 120
Output: 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 purpose
of this problem, assume that your function returns 0 when the reversed
integer 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<int> 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：+,- > <<,>>

Add Two Numbers

You are given two non-empty linked lists representing two non-negative
integers. 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.

You may assume the two numbers do not contain any leading zero, except the
number 0 itself.

Example:

1
2
3

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

代码已经注释的很详细了!
Mycode:

/**
 * 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;
    }
};