[Euler]Problem 36 - Double-base palindromes

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于 2013-04-18 21:11:41 发布

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本文介绍了一个寻找并计算所有小于一百万且在十进制和二进制下均为回文形式的数字之和的问题。通过使用Java编程语言实现了一个简单的暴力破解算法，最终得出这些数字总和为872187，算法运行耗时133毫秒。

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The decimal number, 585 = 1001001001₂ (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

Brute Force！！！

public class Double_BasePalindromes {

	public static void main(String[] args) {
		long before = System.currentTimeMillis();

		new Double_BasePalindromes().calculate();

		System.out.println("elapsed time is : " + (System.currentTimeMillis() - before));
	}

	private void calculate() {
		String baseTwo = "";
		int sum = 0;

		for (int i = 1; i < 1000000; i++) {
			baseTwo = Integer.toBinaryString(i);

			if (isStringPalindromes(String.valueOf(i))) {

				if (isStringPalindromes(baseTwo)) {
					sum += i;
				}

			}

		}

		System.out.println("sum of all numbers is : " + sum);
	}

	private boolean isStringPalindromes(String str) {
		for (int i = 0; i <= str.length() / 2; i++) {

			if (str.charAt(i) != str.charAt(str.length() - 1 - i)) {

				return false;
			}

		}

		return true;
	}

}

console :

sum of all numbers is : 872187
elapsed time is : 133