1.Reverse Integer
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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).
2.Reverse Integer Solution
class Solution {
public:
int reverse(int x) {
int result = 0;
while(x != 0){
result = result * 10 + x % 10;
x = x / 10;
}
return result;
}
};
主要思路就是下面这个可以得到一个整数的所有位数。刚好从最后一位开始得到。
while(x != 0){
int element = x % 10;
x = x / 10;
}
http://www.waitingfy.com/archives/946
本文介绍了一种反转整数的算法实现,通过示例展示如何将整数如123转换为321,并讨论了当输入整数的末位为0时的处理方式以及反转后的整数可能溢出的问题。
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