Leetcode位计算

3095. 或值至少 K 的最短子数组 I 3097. Shortest Subarray With OR at Least K II class Solution: def minimumSubarrayLength(self, nums: List[int], k: int) -> int: n = len(nums) bits = [0] * 30 res = inf def calc(bits): return sum(1 << i for i in range(30) if bits[i] > 0) left = 0 for right in range(n): for i in range(30): bits[i] += (nums[right] >> i) & 1 while left <= right and calc(bits) >= k: res = min(res, right - left + 1) for i in range(30): bits[i] -= (nums[left] >> i) & 1 left += 1 return -1 if res == inf else res 2595. Number of Even and Odd Bits class Solution: def evenOddBit(self, n: int) -> List[int]: res = [0, 0] i = 0 while n: res[i] += n & 1 n >>= 1 i = i ^ 1 return res n >>= 1: Performs a right bitwise shift on ni = i ^ 1: This cleverly toggles the index i between 0 and 1. The ^ operator is the bitwise XOR. Since i is either 0 or 1: If i is 0, 0 ^ 1 is 1. If i is 1, 1 ^ 1 is 0. Binary representation of 8: 1000 Iterations: Iteration 1: n = 8 (1000) n & 1 = 1000 & 0001 = 0000 = 0 res[0] = 0 + 0 = 0 n >>= 1 (n becomes 4, which is 0100) i = 0 ^ 1 = 1 Iteration 2: n = 4 (0100) n & 1 = 0100 & 0001 = 0000 = 0 res[1] = 0 + 0 = 0 n >>= 1 (n becomes 2, which is 0010) i = 1 ^ 1 = 0 Iteration 3: n = 2 (0010) n & 1 = 0010 & 0001 = 0000 = 0 res[0] = 0 + 0 = 0 n >>= 1 (n becomes 1, which is 0001) i = 0 ^ 1 = 1 Iteration 4: n = 1 (0001) n & 1 = 0001 & 0001 = 0001 = 1 res[1] = 0 + 1 = 1 n >>= 1 (n becomes 0, which is 0000) i = 1 ^ 1 = 0 Loop terminates because n is now 0. Return value: res = [0, 1]