解法1:字符串反转 ## 解法一:转换为字符串,要考虑溢出 ## 特殊的案例:< 2**31-1 class Solution: def reverse(self, x: int) -> int: if x>=0: ans = int(str(x)[::-1]) else: ans = -int( str(abs(x))[::-1] ) ## 不用用 -x? ## 考虑溢出 return ans
intCCLeetcode7::CCSolution::reverseMethod1(intx){longres=0;while(x){res=res*10+x%10;x/=1...
classSolution:defreverse(self,x:int)->int:y=int(str(x)[::-1])ifx>=0else-int(str(x)[:0:-1])returnyif-2**31<y<2**31-1else0#作者:jutraman #链接:https://leetcode-cn.com/problems/reverse-integer/solution/pythonzheng-shu-fan-zhuan-by-jutraman/ 表现结果: Runtime: 24 ms, faster...
Reverse Integer 题目来源:https://leetcode.com/problems/reverse-integer/ 问题描述 7. Reverse Integer Easy Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -...Leetcode 7. Reverse Integer Leetcode 7. Reverse Integer 题目...
publicclassSolution{publicintReverse(intx){intresult =0;while(x !=0) {// 取出最后一位数字intdigit = x %10;// 检查溢出if(result >int.MaxValue /10|| (result ==int.MaxValue /10&& digit >int.MaxValue %10)) {return0; }if(result <int.MinValue /10|| ...
class Solution { public: int reverse(int x) { if( overflow(x) == true) { return 0; } int result = 0; while (x!=0) { result = 10*result + x%10; x /= 10; } return result; } private: bool overflow(int x) { if (x / 1000000000 == 0) // x的绝对值小于1000000000, 不...
leetcode Reverse Integer & Reverse a string---重点 https://leetcode.com/problems/reverse-integer/ understanding: 最intuitive的办法就是直接把integer化成string,然后变成list。这里如果化成string,会有溢出的问题,比如integer是1534236469,这个数字反过来就是个很大的数,溢出了,必须返回0. 如果是直接用int计算的,...
publicclassSolution {publicintreverse(intx) {intresult = 0;while(x!=0){ result= result*10+x%10; x/=10; }returnresult; } } Spoiler: 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!
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 2.解法分析 解法需注意: 为零的状况 为负数的状况 溢出怎么处理 结尾一串0怎么处理 代码如下: class Solution { public: int reverse(int x) { // Start typing your C/C++ solution below // DO NO...
17. 18. 19. 20. class Solution: # @return an integer def reverse(self, x): revx = int(str(abs(x))[::-1]) if revx > math.pow(2, 31): return 0 else: return revx * cmp(x, 0) 1. 2. 3. 4. 5. 6. 7. 8.