题目
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
解答
水题，取余，考虑不要越界就可以，我用的这个方法有点蠢，待改进。
class Solution {
public:
int reverse(int x) {
if (x == -2147483648) {
return 0;
}
int xx = abs(x);
vector<int>s;
int flag = int(xx == x);
while (abs(xx) > 0) {
if (s.size() == 0 && xx % 10 == 0) {
xx /= 10;
continue;
}
s.push_back(xx % 10);
xx /= 10;
}
vector<int>::iterator it;
int k = s.size() - 1;
int y = 0;
int danger = s.size() >= 10 ? 1 : 0;
int mz[10] = {6, 4, 6, 3, 8, 4, 7, 4, 1, 2};
int mf[10] = {7, 4, 6, 3, 8, 4, 7, 4, 1, 2};
int m;
for (it = s.begin(); it < s.end(); it++) {
if (flag) {
m = mz[k];
}
else {
m = mf[k];
}
if (danger) {
if (m < *(it)) {
return 0;
}
else if (m > *(it)) {
danger = 0;
}
}
y += *(it) * pow(10, k--);
}
return y * pow(-1, flag + 1);
}
};
题外话
因为申请了一个迭代器，空间复杂度略高。（待续）
