[Leetcode] Reverse Integer

整数反转算法解析

最新推荐文章于 2018-08-12 17:49:21 发布

最新推荐文章于 2018-08-12 17:49:21 发布 · 76 阅读

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本文介绍了一种反转整数的算法实现，通过将整数转换为字符串并逆序处理，再转换回整数形式。讨论了特殊情况如溢出和末尾零的处理，并提供了完整的C++代码实现。

Reverse IntegerDec 26 '116571 / 11753

Reverse digits of an integer.

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

Have you thought about this?

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).

» Solve this problem (link t

尽是做点简单来安慰自己。

#include <stdlib.h>     /* atoi */
#include <stdio.h>

class Solution {
public:
    int countDigit(int n) {
        int c = 0;
        if (n < 0) n=-n;
        while (n > 0) {
            c++;
            n /= 10;
        }
        if (c==0) c = 1;
        return c;
    }
    void itoa(int n, char*a) {
        int len = countDigit(n);
        a[len] = '\0';
        int i = len-1;
        while (n > 0) {
            a[i--] = '0' + n % 10;
            n /= 10;
        }
    } 
    
    int reverse(int x) {
        bool neg = false;
        if (x < 0) {
            neg = true;
            x = -x;
        }
        char a[33], b[33];
        memset(a, 0, 33);
        memset(b, 0, 33);
        itoa(x, a);
        int len = strlen(a);
        char *s = a, *e = a + len - 1;
        while (s < e && *e == '0') e--;
        char *bb = b;
        while (e >= s ) *(bb++) = *(e--); 
        *bb = '\0';
        int y = atoi(b);
        if (neg) return -y;
        else return y;
    }
};