解法1:字符串反转 解法2:数学运算(求余)转换 解法3:递归解法 这个题其实是 LeetCode 9 的基础题。可见,LeetCode 的题目编排就是乱来。 不过呢,相对 LeetCode 9,此题存在一些小的细节必须考虑周全。 题目要求看上去很简单,就是把一个整数从右到左翻过来。比如5,翻过来还是5;123,翻过来就是 321;-124,反过
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0; rev = rev * 10 + pop; } return rev; } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Python python没有溢出问题,处理这题投机取巧 class Solution(object): def reverse(sel...
# for i in range(l): # t=t+res[i]*(10**p) # p=p-1 # if y>0: # return t # else: # return -t defreverse(x): res=0 ifx >0andx <=2**31-1: l=list(str(x)) newl=l[::-1] res=int(''.join(newl)) ifres >2**31: return0 else: returnres elifx <0andx >=...
classSolution{publicintreverse(int x){long res=0;while(x!=0){res=res*10+x%10;x/=10;if(res>Integer.MAX_VALUE||res<Integer.MIN_VALUE)return0;}return(int)res;}}
Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21 Solution: https://leetcode.com/problems/reverse-integer/discuss/229800/Python3-Faster-than-100 ...
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Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 我的代码 #!/bin/env pythonx=raw_input("input a string x:")a='-'ifainx: x=list(x)del x[0]x.reverse()x.insert(0,'-')y="".join(x)print y ...
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse...
When line splitting is on, the colon and modifier letters and integer cannot contain whitespace. I.e., :pr34. If line splitting is off, whitespace is allowed (and ignored).The full list of modifiers is:!: Push the given thing (either a function or a variable) onto the stack, do not...
7 Reverse Integer 原题链接:Reverse Integer 先上代码: classSolution:# @param {integer} x# @return {integer}defreverse(self,x):x=str(x)ifx[0]in['+','-']:x=int(str(x[1:])[::-1])ifx[0]=='+'else-int(str(x[1:])[::-1])else:x=int(str(x)[::-1])if(x>2**31-1)or...