2014
11-19

LeetCode-Reverse Integer[模拟]

Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):

//LeetCode, Reverse Integer
// 时间复杂度O(logn)，空间复杂度O(1)
class Solution {
public:
int reverse (int x) {
int r = 0;

for (; x; x /= 10)
r = r * 10 + x % 10;

return r;
}
};


Java代码:

public class Solution {
public static int reverse(int x){
if(x==Integer.MIN_VALUE)
return Integer.MIN_VALUE;
int ans = 0;
int tmp = Math.abs(x);
while(tmp > 0){
int b = tmp % 10;
ans = ans*10 + b;
tmp /= 10;
}
return x > 0 ? ans:-ans;
}
}

2. #include <stdio.h>
int main()
{
int n,p,t[100]={1};
for(int i=1;i<100;i++)
t =i;
while(scanf("%d",&n)&&n!=0){
if(n==1)
printf("Printing order for 1 pages:nSheet 1, front: Blank, 1n");
else {
if(n%4) p=n/4+1;
else p=n/4;
int q=4*p;
printf("Printing order for %d pages:n",n);
for(int i=0;i<p;i++){
printf("Sheet %d, front: ",i+1);
if(q>n) {printf("Blank, %dn",t[2*i+1]);}
else {printf("%d, %dn",q,t[2*i+1]);}
q–;//打印表前
printf("Sheet %d, back : ",i+1);
if(q>n) {printf("%d, Blankn",t[2*i+2]);}
else {printf("%d, %dn",t[2*i+2],q);}
q–;//打印表后
}
}
}
return 0;
}

3. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

4. 有两个重复的话结果是正确的，但解法不够严谨，后面重复的覆盖掉前面的，由于题目数据限制也比较严，所以能提交通过。已更新算法