2021/05/13
Leetcode:
- medium, 78. Subsets
- medium, 90. subsets II
backtrack
List<List<Integer>> ans = new ArrayList<>();
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums);
backtrack(nums, new ArrayList<>(), 0);
return ans;
}
public void backtrack(int[] nums, List<Integer> cur, int first) {
ans.add(new ArrayList<>(cur));
for(int i = first; i < nums.length; i++) {
if(i != first && i > 0 && nums[i] == nums[i-1])
continue;
cur.add(nums[i]);
backtrack(nums, cur, i+1);
cur.remove(cur.size() -1);
}
}
bitwise
Example: generate subsets of [1,2,3] binary bitmasks: 000 -> [] 001 -> [3] 010 -> [2] 011 -> [2,3] 100 -> [1] 101 -> [1,3] 110 -> [1,2] 111 -> [1,2,3]
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
int n = nums.length;
for(int i = (int)Math.pow(2, n); i < (int)Math.pow(2,n+1); i++) {
// generate 1000 -> substring(1) -> 000
String bitmask = Integer.toBinaryString(i).substring(1);
List<Integer> cur = new ArrayList<>();
for(int j = 0; j < n; j++) {
if(bitmask.charAt(j) == '1')
cur.add(nums[j]);
}
ans.add(cur);
}
return ans;
}
https://leetcode.com/discuss/general-discussion/1073221/All-about-Bitwise-Operations-Beginner-Intermediate
// bitCount
Integer.bitCount(1); // 1
Integer.bitCount(2); // 1
Integer.bitCount(3); // 2
public int bitCount(int n) {
int count = 0;
while(n > 0) {
n = n & (n-1);
count++;
}
return count;
}
More:
- hard, 51. N-Queens