LeetCode 7 Reverse Integer
2017-09-19 09:36
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题目:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
题目链接
题意:
给一个数,将这个数反转过来,假如反转之后对于signed-int而言会溢出的话,那么就返回0,否则返回反转之后的数。
我们可以在int和long long分别设置一个变量,用于存储所给数的绝对值反转之后的数,之后判断这两个临时变量是否相同,假如不相同,则说明存在溢出,否则再判断是否是负数,返回正确答案。
代码如下:
class Solution {
public:
int reverse(int x) {
long long tempLong = 0;
int tempInt = 0, xx = abs(x);
while (xx) {
tempInt = tempInt * 10 + xx % 10;
tempLong = tempLong * 10 + xx % 10;
xx /= 10;
}
if (tempInt != tempLong)
return 0;
if (x < 0) {
return -1 * tempInt;
}
}
};
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
题目链接
题意:
给一个数,将这个数反转过来,假如反转之后对于signed-int而言会溢出的话,那么就返回0,否则返回反转之后的数。
我们可以在int和long long分别设置一个变量,用于存储所给数的绝对值反转之后的数,之后判断这两个临时变量是否相同,假如不相同,则说明存在溢出,否则再判断是否是负数,返回正确答案。
代码如下:
class Solution {
public:
int reverse(int x) {
long long tempLong = 0;
int tempInt = 0, xx = abs(x);
while (xx) {
tempInt = tempInt * 10 + xx % 10;
tempLong = tempLong * 10 + xx % 10;
xx /= 10;
}
if (tempInt != tempLong)
return 0;
if (x < 0) {
return -1 * tempInt;
}
}
};
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