leetcode Single Number II

最新推荐文章于 2018-09-23 00:41:02 发布

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探讨了如何在数组中找到仅出现一次的元素，而其他元素都恰好出现三次的问题。算法采用位操作来实现线性时间复杂度，并且不使用额外内存。通过对比不同实现方式，最终给出了一种清晰易懂的解决方案。

Single Number II

Total Accepted: 4273 Total Submissions: 14218 My Submissions

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?

Similar to I, when the number accumulates for three times, set the bit to 0. I.e., how to simulate a ternary system using a binary system.

I wrote an wrong code like :

class Solution {
 public:
  int singleNumber(int A[], int n) {
    int ones = 0, twos = 0, threes = 0, i, res = 0;
    if (n == 1)
      return A[0];
    for (i = 0; i < n; ++i) {
      res |= A[i];
      threes = twos & A[i];
      res ^= threes;
      
      twos = (twos | (ones & A[i])) ^ threes;
      ones = ones ^ A[i];
    }
    return res;
  }
};

The mistake is that the variable ones needs to be cleared based on the variable threes. At the same time, the variable res seems redundant. A right code is :

class Solution {
 public:
  int singleNumber(int A[], int n) {
    int ones = 0, twos = 0, threes = 0, i, res = 0;
    if (n == 1)
      return A[0];
    for (i = 0; i < n; ++i) {
      //res |= A[i];
      threes = twos & A[i];
      //res ^= threes;
      
      twos = (twos | (ones & A[i])) & (~threes);
      ones = (ones ^ A[i]) & (~threes);
    }
    return ones;
  }
};

However, the code seems not in a standard way. I tried to fix the code in an accessible code:

class Solution {
 public:
  int singleNumber(int A[], int n) {
    int ones = 0, twos = 0, threes = 0, mask = -1, i, res = 0;
    if (n == 1)
      return A[0];
    for (i = 0; i < n; ++i) {
      //res |= A[i];
      threes = twos & A[i];
      //res ^= threes;
      mask = (~threes);
      twos = (twos | (ones & A[i])) & mask;
      mask &= (~twos);
      ones = (ones | A[i] ) & mask;
    }
    return ones;
  }
};

At last, the problem could be changed like this:

Given an array of integers, every element appears m times except for one for n times. Find that the one of n times.

  int findNumber(int A[], int size, int m, int n) {
    int mask = -1, i, j;
	vector<int> times(m+1,0);
	
	for (i = 0; i < size; ++i) {
	  mask = times[0] = -1;
	  times[m] = 0;
	  for (j = m; j >= 1; --j) {
	    times[j] = (times[j] | (times[j-1] & A[i])) & mask;
		mask &= (~times[j]);
	  }
	}
	return times[n];
  }