解析 #include int reverse(int x) { int y=0; while(x!=0) { y=y*10+x%10; x=x/10; } return y; } int main() { int a; printf("请输入一个正整数:"); scanf("%d",&a); printf("%d的逆序数是%d\n",a,reverse(a)); return 0; } ...
public: int reverse(int x) { int y = 0; int n; while (x != 0) { n = x % 10; if (y > INT_MAX / 10 || y < INT_MIN / 10) { return 0; } y = y * 10 + n; x /= 10; } return y; } }; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15....
【青鸟飞扬教育】整数反转 正常情况下,此题使用int肯定会溢出的,所以需要在循环中提前判断一下是否超过[-2^31, 2^31 - 1]。 class Solution { public int reverse(int x) { int res = 0; while (x != 0) { int temp = x % 10; x = x / 10; if ((res > 0) && (res > (Integer.MAX_...
using namespace std; int reverse(int x) { int i=0,f=0;char str[100]; if(x>0) { f=1; }else { f=0;x=-x;} if (x==0) return 0; while(x>0) { str[i]=x%10+'0'; x/=10; i++; } str[i]='\0'; if(f==1) return atoi(str); if(f==0) return -atoi(str); ...
classSolution1{publicintreverse(intx){while(x %10==0) { x /=10; }booleantag=false;if(x <0) { tag =true; x = x * -1; }Strings=String.valueOf(x);char[] ch = s.toCharArray();for(inti=0;i < ch.length;i++) {intk=ch.length - i -1;if(k < ch.length /2)break;char...
class Solution { public static int reverse(int x) { boolean positive = true; char[] a = String.valueOf(x).toCharArray(); if ((int) a[0] == 45) { positive = false; a = Arrays.copyOfRange(a, 1, a.length); } System.out.println(Arrays.toString(a)); for (int i = 0; i ...
int reverse(int x) { int result = 0; while(x != 0){ result = result * 10 + x % 10; x = x / 10; } return result; } }; 主要思路就是下面这个可以得到一个整数的所有位数。刚好从最后一位开始得到。 while(x != 0){ int element = x % 10; ...
int reverse(int n){ int r=0; while(n!=0) { r*=10; r+=(n%10); n/=10; } return r;}
int变成string，string变成chararray，chararray倒序遍历变回string string变回int 记得判断正负。也可以用/10的余数取数字然后再乘10加回来 例如：public String reverseSting(String inputString) { char[] inputStringArray = inputString.toCharArray();String reverseString = "";for (int i = input...
Reverse Int Task Given: number, your task is to implement function that reverse digits of this number. For example: reverse(123); // 321 reverse(233); // 332 reverse(535); // 535 reverse(95034); // 43059 Write your code in `src/index.js. All test cases are designed as “error-...