一：Reverse Integer
题目：
Reverse digits of an integer.
Example1:x = 123, return 321Example2:x = -123, return -321
链接：https://leetcode.com/problems/reverse-integer/
分析：这题通过不断取余将余数存放在一个vector中，然后乘以相应的10^i次方相加即可，这里主要考虑是否overflow,因此将result设为long long int，当大于int的最大值0x7fffffff或者小于int的最小值0x800000000时，则返回0，，否则返回result。
代码：
#define maxInt (int)0x7fffffff// int中最大值==2147483647unsigned int 最大值为4294967295#define minInt (int)0x80000000// int中最小值2147483648class Solution {public:int reverse(int x) {vector<int> v;while(x != 0){int a = x%10;v.push_back(a);x = x/10;}long long int result = 0;for(int i = 0; i < v.size(); i++){result += v[i]*pow(10.0,v.size()- i – 1);}if(result > maxInt || result < minInt) return 0;// 通过long long int来解决溢出问题return result;}};二：String to Integer (atoi)
题目：Implementatoito convert a string to an integer.
链接：https://leetcode.com/problems/string-to-integer-atoi/
分析：这道题不难，关键是满足的测试样例比较变态，比如“ 123456ab78”，也要返回123456等等
代码：
#define Max (int)0x7fffffff#define Min (int)0x80000000class Solution {public:int atoi(string str) {if(str.size() <= 0) return 0;for(int i = 0; i < str.size(); i++){ // 去除前面的空格if(str[i] != ' '){str = str.substr(i);break;}}int c = 1;int len = str.size();if(isdigit(str[0])) {// 如果第一个为数字 表示整型 则前面放一个+号str = "+" + str;len = len +1;}if(str[0] == '+') c = 1;else if(str[0] == '-')c = -1;else return 0;for(int i = 1; i < len; i++){ //去除后面字母开头的所有字符if(!isdigit(str[i]) || i >= 12) { //超过了12位的不要了 否则有可能会超过long long 的范围str = str.substr(0, i);break;}}len = str.size();long long int result = 0;;for(int i = 1; i < len; i++){if(!isdigit(str[i]))return 0;stringstream ss;int temp;ss << str[i];ss >> temp;result += temp * pow(10.0, len-i-1.0)*c;}if(result > Max) return Max;if(result < Min) return Min;return result;}};三：Palindrome Number
题目：
Determine whether an integer is a palindrome. Do this without extra space.
Some hints:
Could negative integers be palindromes? (ie, -1)
If you are thinking of converting the integer to string, note the restriction of using extra space.
You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
There is a more generic way of solving this problem.
链接：
https://leetcode.com/problems/palindrome-number/
分析：这题主要有几个限制：1不能用extra space,所以不能转换成string再求回文 2：先求reverse integer，那么溢出怎么办？ 所以这题是先求出int的位数，然后分别取出最低位和最高位进行比较。
class Solution {public:bool isPalindrome(int x) {if(x < 0) return false;int d = 0;int y = x;while(y != 0){ // 先将x的位数求出来y = y/10;d++;}while(x != 0){int lowBit = x % 10; // 求出x的最低位int highBit = x / pow(10.0, d-1.0); // 求出x的最高位if(lowBit != highBit) return false;x = (x – lowBit – highBit * pow(10.0, d-1.0))/10;d = d – 2;}return true;}};
奢侈地享受旅不问人，行随己意的潇洒。
