每周一道算法题001:回文数

题目:

找出大于10的最小的2进制,8进制,10进制都是回文数的最小的数。回文数指的是正读和反读都是一样的数,例如:33,10001,123454321...

思路:

先转换进制,然后统一处理成字符串进行比较

解答:

PHP

function execute(){
    $x = 11;
    while (1) {
        if ($x == strrev($x)
            && decbin($x) == strrev(decbin($x))
            && decoct($x) == strrev(decoct($x))) {
            break;
        }
        $x += 2;
    }
    return $x;
}

$result = execute();
echo $result;

golang

package main

import (
    "fmt"
    "strconv"
)

func main() {
    result := Execute()
    fmt.Println(result)
}

func Execute() string {
    xStr := ""
    for x := 11; ; x += 2 {
        xStr = strconv.Itoa(x)
        xBin := strconv.FormatInt(int64(x), 2)
        xOct := strconv.FormatInt(int64(x), 8)
        if xStr == Reverse(xStr) && xBin == Reverse(xBin) && xOct == Reverse(xOct) {
            break
        }
    }
    return xStr
}

// 字符串翻转
func Reverse(s string) string {
    runes := []rune(s)
    for from, to := 0, len(runes)-1; from < to; from, to = from+1, to-1 {
        runes[from], runes[to] = runes[to], runes[from]
    }
    return string(runes)
}
上一篇:每周一道算法题002:四则运算


下一篇:java中线程机制