Single Number II

最新推荐文章于 2022-07-02 13:41:47 发布

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本文介绍了一种算法挑战：在一个每个元素都出现三次的整数数组中找到仅出现一次的数字。提供了三种解决方案，包括使用哈希表、位操作及逐位统计的方法，并强调了位操作方案在不使用额外内存的情况下实现线性时间复杂度的优势。

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

    public int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        Hashtable<Integer, Integer> table = new Hashtable<Integer, Integer>();
        for(int i = 0; i < A.length; i++){
            if(table.containsKey(A[i]))
                table.put(A[i], table.get(A[i]) + 1);
            else
                table.put(A[i], 1);
        }
        for(int i = 0; i < A.length; i++){
            if(table.get(A[i]) != 3)
                return A[i];
        }
        return 0;
    }

Hash uses extra space.

Bit operation does not need extra space.

	public int singleNumber2(int[] A){
		int ones = 0, twos = 0;
		int commonBitMask = 0;
		for(int i = 0; i < A.length; i++){
			twos |= (ones & A[i]);
			ones ^= A[i];
			commonBitMask = ~(ones & twos);
			ones &= commonBitMask;
			twos &= commonBitMask;
		}
		return ones;
	}

非常牛逼的解法

	public int singleNumber3(int A[]) {
		int[] count = new int[32];
		int result = 0;
		for (int i = 0; i < 32; ++i) {
			count[i] = 0;
			for (int j = 0; j < A.length; ++j) {
				if (((A[j] >> i) & 1) != 0)
					count[i] = (count[i] + 1) % 3;
			}
			result |= (count[i] << i);
		}
		return result;
	}