一、問題描述

功能描述：對一個二進位制數的每位進行0和1反轉，求翻轉後的二進位制所對應的十進位制
輸入：110
輸出：1

要求實現方法：
public String getNumber(BigInteger data)
{
    //TODO
    return "";
}

二、演算法

public String getNumber(BigInteger data) {
		String biStr = String.valueOf(data);
		char[] arr = biStr.toCharArray();
		for (int i = 0; i < arr.length; i++) {
			if (arr[i] == '0') {
				arr[i] = '1';
			} else {
				arr[i] = '0';
			}
		}
		
		//reverse.
		int result = 0;
		for (int i = 0; i < arr.length; i++) {
			int temp = 1;
			int isum = 0;
			if(Character.getNumericValue(arr[i]) != 0) {
				if(i == arr.length -1 ) {
					result += 1;// 2的零字幕 = 1
					continue;
				}
				while (temp <= i) {
					isum = temp * 2;
					temp ++;
				}
			}
			result += isum;
		}
		return String.valueOf(result);
	}
 

package com.albertshao.csi.interview;

import java.math.BigInteger;

/**
 * @author albertshao
 */
public class Main17 {

	public static void main(String[] args) {

		Main17 m = new Main17();
		System.out.println(m.getNumber(new BigInteger("110")));
	}

	
	public String getNumber(BigInteger data) {
		String biStr = String.valueOf(data);
		char[] arr = biStr.toCharArray();
		for (int i = 0; i < arr.length; i++) {
			if (arr[i] == '0') {
				arr[i] = '1';
			} else {
				arr[i] = '0';
			}
		}
		
		//reverse.
		int result = 0;
		for (int i = 0; i < arr.length; i++) {
			int temp = 1;
			int isum = 0;
			if(Character.getNumericValue(arr[i]) != 0) {
				if(i == arr.length -1 ) {
					result += 1;// 2的零字幕 = 1
					continue;
				}
				while (temp <= i) {
					isum = temp * 2;
					temp ++;
				}
			}
			result += isum;
		}
		return String.valueOf(result);
	}
}
 

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