Leetcode_191_Number of 1 Bits

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Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming
weight).

For example, the 32-bit integer ’11' has binary representation

00000000000000000000000000001011

,
so the function should return 3.

思路：

(1)题意为给定一个无符号数，返回其对应二进制数中1的个数。

(2)该题主要考察进制的转换操作。java中没有无符号数字(为什么没有可以百度下)。无符号数即给定的数都是非负数。这样我们可以通过java自带类库中的Integer.toBinaryString(value)将指定整数专为二进制数，然后求得二进制数中包含1的个数即可。详情见下方代码。

(3)希望本文对你有所帮助。

算法代码实现如下:

/**
	 * @param liqq 直接用类库
	 */
	public int hammingWeight(int value) {
		int count = 0;
		String binaryString = Integer.toBinaryString(value);
		for (int i = 0; i < binaryString.length(); i++) {
			char charAt = binaryString.charAt(i);
			if (charAt == '1') {
				count++;
			}
		}
		return count;
	}

/**
	 * @param liqq 简化类库中toBinaryString方法的使用
	 */
	// you need to treat n as an unsigned value
	public int hammingWeight(int value) {
		int count = 0;
		// String binaryString = Integer.toBinaryString(value);
		char[] digits = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
				'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l',
				'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x',
				'y', 'z' };
		char[] buf = new char[32];
		int charPos = 32;
		int radix = 2;
		int mask = 1;
		do {
			buf[--charPos] = digits[value & 1];
			value >>>= 1;
		} while (value != 0);

		String binaryString = new String(buf, charPos, (32 - charPos));

		for (int i = 0; i < binaryString.length(); i++) {
			char charAt = binaryString.charAt(i);
			if (charAt == '1') {
				count++;
			}
		}

		return count;
	}