位操作应用

1. 给定一个含不同整数的集合，返回其所有的子集

S = {1,2,3}


0 000 {}
1 001 {1}
2 010 {2}
3 011 {1,2}
4 100 {3}
5 101 {1,3}
6 110 {2,3}
7 111 {1,2,3}

class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public List<ArrayList<Integer>> subsets(int[] nums) {
        List<ArrayList<Integer>> result = new ArrayList<List<Integer>>();
        int n = nums.length;
        Arrays.sort(nums);
        
        // 1 << n is 2^n
        // each subset equals to an binary integer between 0 .. 2^n - 1
        // 0 -> 000 -> []
        // 1 -> 001 -> [1]
        // 2 -> 010 -> [2]
        // ..
        // 7 -> 111 -> [1,2,3]
        for (int i = 0; i < (1 << n); i++) {
            List<Integer> subset = new ArrayList<Integer>();
            for (int j = 0; j < n; j++) {
                // check whether the jth digit in i's binary representation is 1
                if ((i & (1 << j)) != 0) {
                    subset.add(nums[j]);
                }
            }
            result.add(subset);
        }
        
        return result;
    }
}

2. ven a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

// Maximum Product of Word Lengths
// Time Complexity: O(n^2), Space Complexity: O(n)
public class Solution {
    public int maxProduct(String[] words) {
        final int n = words.length;
        final int[] hashset = new int[n];

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < words[i].length(); ++j) {
                hashset[i] |= 1 << (words[i].charAt(j) - 'a');
            }
        }

        int result = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                int tmp = words[i].length() * words[j].length();
                if ((hashset[i] & hashset[j]) == 0 && tmp > result) {
                    result = tmp;
                }
            }
        }
        return result;
    }
}

3. Integer Replacement

Given a positive integer n and you can do operations as follow:

1. If n is even, replace n with n/2.
2. If n is odd, you can replace n with either n + 1 or n - 1.

class Solution {
public:
    int integerReplacement(int n) {
		int count = 0;
        if(n == 0x7fffffff) return 32;
		while (n > 1) {
			
			if ((n & 1) == 0) n >>= 1;
			else if ((n & 0b11) == 0b11 && (n!=3)) n++; 
			else n--;
			count++;
		}
		return count;
    }
};