有一個超大超大的10進位數字字串,如何將其轉換成二進位字串?
由於數字特別大,沒辦法轉換成number放在變數裡toString(2)
依照十進制轉二進制的原理,自己模擬除2取餘
一種自己寫演算法 應該不會太難 另一種 交給後台處理
private static String process(ArrayList<String> imp,String HexStr){ String rem = ""; Long remaind = (long) 0; ArrayList<String> newList = new ArrayList<String>(); Iterator<String> itor = imp.iterator(); while(itor.hasNext()){ String ele = itor.next(); Long num = Long.parseLong(rem+ele); Long quot = num / 16; remaind = num%16; rem = remaind.toString(); if(quot > 0){ newList.add(padString(quot.toString(),16,"0",0)); } }; String hexRem = Long.toHexString(remaind); if(newList.isEmpty()){ return hexRem; } else{ return process(newList,hexRem)+hexRem; } } public static String D2Hex(String dNum){ String tmp = dNum; ArrayList<String> splitNum = new ArrayList<String>(); while(tmp.length() > 16){ String high = tmp.substring(0,tmp.length() - 16); splitNum.add(high); String low = tmp.substring(tmp.length() - 16); tmp = low; } splitNum.add(tmp); return process(splitNum,""); } public static String D2B(String dNum){ String hex = D2Hex(dNum); StringBuilder bin= new StringBuilder(); for(int i =0; i < hex.length(); i ++){ String tmp = Integer.toBinaryString( Integer.parseInt(hex.substring(i,i+1),16)); bin.append(padString(tmp,4,"0",0)); } return bin.toString(); }
超大數結果未驗證 0------>0--->0000135,9234------>14bd82--->0001010010111101100000109999,9999,9999 -->2386f26fc0ffff--->001000111000011011110010011011111100000011111111111111111,0000,11111111111111. 11100001101111001001101111110000010000000000000009,9999,9999,9999,9999------>16345785d89ffff --->000101100011010001010111100001011101100010011111111111111111
隨便寫一張吧
function toBin(str) { var arr = []; var remainder,i,str2,num,char; while(str.length>0) { str2 = "";remainder=0; for(i=0;i<str.length;i++) { // str2 = str组成的十进制数 / 2 num = str.charCodeAt(i)-0x30; // num to String num = remainder*10 + num; char = Math.floor(num/2).toString(); // 忽略最高为的0 , 即最高为如果是 0 则不放入 str2 if(!(char === "0" && str2 === "")) { str2 += char;} remainder = num%2; } str = str2; arr.push(remainder); // 保存余数 } return arr.reverse().join(''); } console.log(toBin("3")); // 11 console.log(toBin("9")); // 1001 console.log(toBin("10")); // 1010 console.log(toBin("120")); // 1111000
我想問,超大,有多大
如果只是顯示的話, 可以考慮讓後台轉
依照十進制轉二進制的原理,自己模擬除2取餘
一種自己寫演算法 應該不會太難 另一種 交給後台處理
超大數結果未驗證
9999,9999,9999,99999999,08949, cef80ffffffff--->0100111011100010110101101101010000010101101110000101101011001110111110000000111111111111111111111111111111111111111111111111111111111111111111111111111111111111 🎜0------>0--->0000
135,9234------>14bd82--->000101001011110110000010
9999,9999,9999 -->2386f26fc0ffff--->00100011100001101111001001101111110000001111111111111111
1,0000,11111111111111
. 1110000110111100100110111111000001000000000000000
9,9999,9999,9999,9999------>16345785d89ffff --->000101100011010001010111100001011101100010011111111111111111
隨便寫一張吧
我想問,超大,有多大
如果只是顯示的話, 可以考慮讓後台轉