leetcode(2)--整数反转

    技术2025-09-24  40

    class Solution { public:     int reverse(int x) {         int y = 0;     while(x!=0)     {         if(y>214748364  || y < -214748364 )         {return 0;}         y = y*10+x%10;         x=x/10;     }     return y;     }      };

    Processed: 0.012, SQL: 9