leetcode — reverse-integer】的更多相关文章

Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 click to show spoilers. SOLUTION 1: 注意越界后返回0.先用long来计算,然后,把越界的处理掉. public class Solution { public int reverse(int x) { long ret = 0; while (x !…
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought throu…
第一天上课,数据库老师说对于计算机系的学生,凡是在课本上学到的专业知识都是过时的.深以为然,感觉大学两年半真的不知道学了什么,为未来感到担忧,C++也不敢说是精通,入门还差不多.最近丧的不行,不管怎么样,每天还是要进步一点点. 题目: Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 感觉不难,写了才发现自己没考虑溢出的问题,为判断溢出折腾了好久,心好累,提…
Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought throug…
问题: Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return –321   Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! If the in…
题目: Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! If the integ…
1.在进入while之前,保证x是非负的: 2.符号还是专门用flag保存 =================== 3.另一思路:将integer转换成string,然后首位swap,直至中间: class Solution: # @return an integer def reverse(self, x): ret = 0 flag = 1 if x < 0: flag = -1 x *= -1 while(x!=0): ret = ret*10+x%10 x = x/10 return r…
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! If the integer'…
  核心思想:原数对10取余数赋值给新数后降一位,再把新数升一位加上下一次原数取余值,直到原数降为0. 解法如下: int reverse(int x) { bool minus = false; ) { x= -x; minus = true; } ; while(x) { a*=; a+=x%; x/=; } return minus?-a:a;…
public class Solution { public int reverse(int x) { int ret=0; while(x!=0) { int t=x%10; ret=ret*10+t; x=x/10; } return ret; }} public class Solution { public boolean isPalindrome(int x) { if(x<0) return false; if(x==0) return true; int ret=0; int xo…