蓝桥杯基础-015-Hexadecimal_To_Decimal

题意

输入一个字符串(16进制数) ( 8 位)

输出一个十进制数 ( 要用long 由于Max（abs（INT）） = 2147483648 -(E)> 2^31 -(16)> 7F_FFF_FFF , long -> 2^63 -> 7FF_FFF_FFF_FFF_FFF)

代码

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		char[] HexAlphabet = new char[16];
		char[] A_F_Alphabet = new char[6];
		for (int i = 0; i < A_F_Alphabet.length; i++) {
			A_F_Alphabet[i] = (char)((int)'A' + i);
		}
		for (int i = 0; i < HexAlphabet.length; i++) {
			if( i <= 9)
				HexAlphabet[i] = (char)((int)'0' + i);
			else
				HexAlphabet[i] = A_F_Alphabet[ i%10 ];
		}
		Scanner keyin = new Scanner(System.in);
		char[] hex_input;
		String str_in = keyin.nextLine();
		hex_input = str_in.toCharArray();
		long digit_power = 1;
		long result = 0;
		for (int i = str_in.length() - 1; i >= 0; i--) {
			result += (long)(binarySearch(HexAlphabet,hex_input[i])*digit_power);
			digit_power *= 16;
		}
		System.out.println(result);

	}
	public static int binarySearch(char[] array,char target){
		int low = 0,high = array.length - 1 ,mid;
		do{
			mid = (low + high)/2;
			if( target > array[mid] ){
				low = mid + 1;
			}
			else if( target < array[mid] ){
				high = mid - 1;
			}
			else
				return mid;
		}while(low <= high);
		return -1;
	}

}

Summary

1.算清楚输入和输出范围,防止溢出

2.思路

i) 建立HexAlphabet对映表0 <-> 15 -> '0' + '9' <-> 'A'  <-> 'F' 即 下标与16进制字符一一对映 && （int）‘9’ < （int）'A'

ii)binarySearch + HexAlphabet + digit_power (为每一位上的权值,初始为1,随字符遍历 digit_power *= 16)  -> result : long += binarySearch(…)

iii)输出result