[LeetCode] 137. Single Number II 单独数 II
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Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,3,2] Output: 3
Example 2:
Input: [0,1,0,1,0,1,99] Output: 99
解法:参考
Java:
public int singleNumber(int[] A) { int ones = 0, twos = 0; for(int i = 0; i < A.length; i++){ ones = (ones ^ A[i]) & ~twos; twos = (twos ^ A[i]) & ~ones; } return ones; }
Python:
class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int """ x1, x2, mask = 0, 0, 0 for i in nums: x2 ^= x1 & i; x1 ^= i; mask = ~(x1 & x2); x2 &= mask; x1 &= mask; return x1
C++:
class Solution { public: int singleNumber(vector<int>& nums) { int one = 0, two = 0, three = 0; for (int i = 0; i < nums.size(); ++i) { two |= one & nums[i]; one ^= nums[i]; three = one & two; one &= ~three; two &= ~three; } return one; } };
C++:
class Solution { public: int singleNumber(vector<int>& nums) { int a = 0, b = 0; for (int i = 0; i < nums.size(); ++i) { b = (b ^ nums[i]) & ~a; a = (a ^ nums[i]) & ~b; } return b; } };
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