[Leetcode] 7 - Reverse Integer
2015-01-07 16:28
267 查看
原题链接: https://oj.leetcode.com/problems/reverse-integer/
做法直接看代码,无需对负数做特殊处理。但是必须考虑溢出的情况,考虑溢出的判断则将res * 10 + num % 10 < INT_MAX的等式转化为res < (INT_MAX - num % 10) / 10,否则res * 10 + num % 10是溢出的从而导致判断错误。
class Solution {
public:
int reverse(int x) {
int num = x;
int res = 0;
while (num) {
if (x > 0) {
if (res > (INT_MAX - num % 10) / 10) {
return 0;
}
} else if (x < 0){
if (res < (INT_MIN - num % 10) / 10) {
return 0;
}
}
res = res * 10 + num % 10;
num /= 10;
}
return res;
}
};
做法直接看代码,无需对负数做特殊处理。但是必须考虑溢出的情况,考虑溢出的判断则将res * 10 + num % 10 < INT_MAX的等式转化为res < (INT_MAX - num % 10) / 10,否则res * 10 + num % 10是溢出的从而导致判断错误。
class Solution {
public:
int reverse(int x) {
int num = x;
int res = 0;
while (num) {
if (x > 0) {
if (res > (INT_MAX - num % 10) / 10) {
return 0;
}
} else if (x < 0){
if (res < (INT_MIN - num % 10) / 10) {
return 0;
}
}
res = res * 10 + num % 10;
num /= 10;
}
return res;
}
};
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