LeetCode Reverse Integer 带考虑第二种特殊情况

Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

很简单的题目，但是如果要考虑到特殊情况的话，还是比较不容易的，下面是accepted的最简单的程序。

int reverse(int x) {  int result = 0;  while (x)  {  result = result*10 + x%10;  x /= 10;  }  return result;  }  

考虑到第二个特殊情况，下面程序也是accepted的，不过好像LeetCode并不测试这个特殊情况。

所以我这里也只是小小处理了一下：reverse之后如果上溢，返回最大值INT_MAX； 如果是下溢，返回最小值INT_MIN

int reverse(int x) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.if(-10<x && x<10 || x == INT_MAX || x == INT_MIN) return x;int xSave = x;vector<int> vi;vector<int> viMax;vector<int> viMin;bool signX = false;if(x<0) signX = true;int maxX = INT_MAX;int minX = INT_MIN;while (x!=0){vi.push_back(x % 10);x /= 10;if(!signX){viMax.push_back(maxX % 10);maxX /= 10;}else{viMin.push_back(minX % 10);minX /= 10;}}bool upOverInt = false;bool downOverInt = false;if(!signX && maxX == 0){for(auto iter1 = vi.begin(); iter1 != vi.end();  iter1++){upOverInt = false;auto iter2 = viMax.end();iter2--;if(*iter1 > *iter2){upOverInt = true;break;}if(*iter1 == *iter2){upOverInt = true;}if(!upOverInt) break;} }else if(signX && minX == 0){for(auto iter1 = vi.begin(); iter1 != vi.end();  iter1++){downOverInt = false;auto iter2 = viMin.end();iter2--;if(*iter1 > *iter2){downOverInt = true;break;}if(*iter1 == *iter2){downOverInt = true;}if(!downOverInt) break;} }if(upOverInt) return INT_MAX;if(downOverInt) return INT_MIN;int i = 0;int xtemp = 0;for(auto iter = vi.begin(); iter != vi.end(); iter++){xtemp = xtemp*10 + *iter;}return xtemp;}

总结：
程序居然长了那么多。不得不感叹要编写“完美”的程序是多么艰难啊。看来够用就好，有时候也是不错的选择！