class Solution: def reverseLeftWords(self, s: str, n: int) –> str: s = list(s) s[0:n] = list(reversed(s[0:n])) s[n:] = list(reversed(s[n:])) s.reverse() return "".join(s) 总结 今天的内容就结束了,继续加油呀! 神龙|纯净稳定代理IP免费测试>>>天启|企业级代理...
publicclassSolution {publicintreverse(intx) {intres = 0;while(x != 0) {if(res > Integer.MAX_VALUE / 10 || res < Integer.MIN_VALUE / 10)return0; res= res * 10 + x % 10; x/= 10; }returnres; } } Reference: https://leetcode.com/discuss/39594/golfing-in-python...
Method 1: Reverse a Number in Python using While Loop In this approach, we will divide the number by 10 and add the remainder to the sum. Program/Source Code Here is the source code of the Python Program to reverse a given number. n=int(input("Enter number: "))rev=0while(n>0): ...
class Solution { public: int reverse(int x) { long long result = 0; while...C#代码: public class Solution { public int Reverse(int x) { long result = 0; while...Python参考代码: (注意Python取余的方式和类C语言不通,所以要分正数和负数情况不同对待) class Solution: # @return an intege...
>>> from miasm2.expression.expression import ExprInt32 >>> sb.symbols[machine.mn.regs.ECX] = ExprInt32(-3) >>> symbolic_pc = sb.emul_ir_blocs(ira, 0, step=True) ___ ECX 0x1 ___ ECX 0x1 EBX (EBX_init+0x
This Blog provides a comprehensive guide to creating prime numbers, perfect numbers, and reverse numbers in Python. Learn More about Python Numbers!
>>> from miasm2.expression.expression import ExprInt32 >>> sb.symbols[machine.mn.regs.ECX] = ExprInt32(-3) >>> symbolic_pc = sb.emul_ir_blocs(ira, 0, step=True) ___ ECX 0x1 ___ ECX 0x1 EBX (EBX_init+0x
python3代码: 1classSolution:2defreverseString(self, s: List[str]) ->None:3"""4Do not return anything, modify s in-place instead.5"""6#s.reverse()7#for i in range(len(s) // 2):8#s[i], s[-1-i] = s[-1-i], s[i]9i, j = 0, len(s) - 110whilei <j:11s[i], ...
这是非常简单的一道题,唯一需要注意的就是越界问题。 int类型的范围是:-2 ^31 ~ 2 ^31 - 1(-2147483648~2147483647),假设我们输入的整数是1234567899,reverse后就变成了9987654321,超出int最大范围,也就会出现越界错误。所以为了避免这种情况,我们在定义最终返回结果的时候,应该使用long型而不是int型。
int main() { std::vector<int> nums {1, 2, 3, 4, 5, 6}; std::reverse(nums.begin(), nums.end()); std::cout << "逆转后的容器元素:"; for(auto num : nums) { std::cout << num << " "; } std::cout << std::endl; return 0; } ``` 以上示例中,我们创建了一个包含数...