Easy_02

题目

难点：如何判断翻转之后是否超出了int类型的范围

解决办法

(1) 使用long int 类型 扩大范围进行判断，最后进行强制类型转换返回int类型

(2) 将范围的各位数 存储在vector数组中注意比较。不好，消耗内存的同时需要额外的循环比较判断。（做题时采用了该方法，结果并不理想）

int reverse(int x){

int c = 1.0;

if (x < 0){

c = -1.0;

}

int result = 0;

vector<int> remains; //余数部分

int max[10] = {2,1,4,7,4,8,3,6,4,8};

int min[10] = {2,1,4,7,4,8,3,6,4,7};

int k = 0;

int remain;

while(x != 0){

remain = x % 10 * c;

remains.push_back(remain);

x = x / 10;

if (k > 10)

return result;

++k;

}

if (k == 10){

if (c == -1.0){

for (int i = 0; i < 10;++i){

if (remains[i] < min[i])

break;

else if (remains[i] == min[i])

    continue;

else

return 0;

}

}

else

for (int i = 0; i < 10; ++i){

if (remains[i] < max[i])

break;

else if (remains[i] == max[i])

    continue;

else

return 0;

}

}

int step = pow(10,remains.size() - 1);

for (vector<int>::iterator it = remains.begin();it != remains.end();++it){

result = result + *it * step;

step /= 10;

}

return result * c ;

}