高效計算一個字節中所有的1的個數。

問題：高效計算一個字節中所有的1的個數

思路1：轉成字符串再輸出

這種方式效果肯定不好，因爲字符串效率低，所以不推薦（代碼略）

思路2：利用位的與計算

基本方法是對每一位進行與運行，如果是不是0，計數器就加1，這種效率還可以，代碼如下：

public static int getOnes(byte b) {
	int count =  (b & 0b00000001) == 0 ? 0 : 1;
	count += (b & 0b00000010) == 0 ? 0 : 1;
	count += (b & 0b00000100) == 0 ? 0 : 1;
	count += (b & 0b00001000) == 0 ? 0 : 1;
	count += (b & 0b00010000) == 0 ? 0 : 1;
	count += (b & 0b00100000) == 0 ? 0 : 1;
	count += (b & 0b01000000) == 0 ? 0 : 1;
	count += (b & 0b10000000) == 0 ? 0 : 1;		
	return count;
}

實際上，代碼還可以進一步優化，可以把變量count給省略，代碼如下所示：

public static int getOnes(byte b) {
	return ((b & 0b00000001) == 0 ? 0 : 1) + 
			 ((b & 0b00000010) == 0 ? 0 : 1) + 
			 ((b & 0b00000100) == 0 ? 0 : 1) + 
			 ((b & 0b00001000) == 0 ? 0 : 1) + 
			 ((b & 0b00010000) == 0 ? 0 : 1) + 
			 ((b & 0b00100000) == 0 ? 0 : 1) + 
			 ((b & 0b01000000) == 0 ? 0 : 1) + 
			 ((b & 0b10000000) == 0 ? 0 : 1); 
}

實際上以上的代碼的最外層括號是可以去掉的，但是了加上更清楚，而且在編輯的時候也會編譯器也會去掉。

方案3：查表法

實際上一個字節只有256個值，如果我們建立一個數組，可以直接通過查表法直接找出結果。這是算法設計與分析中的經典的空間換時間的思路，這樣的效率是最高的，數組如下所示。在調用的時候需要注意一點，如果直接使用一個字節訪問數組，如 byte b = 128; 會報錯。因爲字節在訪問時會被轉換成 int 型，而在轉換時由於高位的表示問題，當 b 大於127時，會變成負數。所以正確的訪問方式是：ones[b&0xff]。

static int[] ones = {
        0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 
        1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
        1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
        1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
        2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 
        3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
        3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 
        4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
        };

附：測試代碼

	public static void main(String[] args) {
		for (int i = 0; i < 256; i++) {
			bts[i] = getOnes((byte) i);
			System.out.print(bts[i] + ", ");
		}
		System.out.println();
		for (int i = 0; i < 256; i++) {
			System.out.print(ones[i] + ", ");
		}
	}

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,