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我想實現代碼將十進制轉換爲具有一定精度的二進制文件,因爲我使用堆棧和鏈接列表來添加計算的非十進制和小數部分。然後我使用Stringbuilder逐個彈出/輪詢元素以生成最終的二進制數。來源:http://www.geeksforgeeks.org/convert-decimal-fraction-binary-number/從堆棧和鏈接列表JAVA中缺少數字?
當我將元素推入堆棧/列表時,我發現他們正在被推送(使用o/p stmts)。出於某種原因,我在彈出元素時看不到它們。
這裏是我的代碼
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
公共類BinaryToDecimal {
public String toBinary(float n, int p){
int non_dec = (int) Math.floor(n);
Stack<Integer> s_non_dec = new Stack<>();
LinkedList<Integer> q_dec = new LinkedList<>();
float dec = n - non_dec;
int quotient = 1;
while(quotient > 0){
quotient = non_dec/2;
int remainder = non_dec%2;
System.out.println("quotient"+quotient+"non_dec"+non_dec+"remainder"+remainder);
s_non_dec.push(remainder);
non_dec = quotient;
}
while(p>0){
System.out.println("before dec"+dec);
dec = dec*2;
System.out.println("after dec"+dec);
if(dec >=1){
System.out.println("add 1");
q_dec.add(1);
dec = dec - 1;
}
else{
System.out.println("add 0");
q_dec.add(0);
}
p--;
}
StringBuilder sb = new StringBuilder();
for(int i=0;i<s_non_dec.size();i++){
System.out.println("pop"+s_non_dec.peek());
sb.append(s_non_dec.pop());
}
sb.append('.');
for(int i=0;i<q_dec.size();i++){
System.out.println("poll"+q_dec.peek());
sb.append(q_dec.poll());
}
return sb.toString();
}
public static void main (String args[]){
BinaryToDecimal btd = new BinaryToDecimal();
System.out.println(btd.toBinary(2.47f, 5));
}
}
我的輸出:
quotient1non_dec2remainder0
quotient0non_dec1remainder1
before dec0.47000003
after dec0.94000006
add 0
before dec0.94000006
after dec1.8800001
add 1
before dec0.8800001
after dec1.7600002
add 1
before dec0.7600002
after dec1.5200005
add 1
before dec0.52000046
after dec1.0400009
add 1
pop1
poll0
poll1
poll1
1.011
由上述可見,甚至強硬。我推1和0進入我的堆棧,我的輸出只有1爲非小數部分,而不是1和0!小數部分也是一樣! 我一直在看這個代碼幾個小時,任何幫助表示讚賞!