[LeetCode] Single Number II

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Single Number II

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?

解题思路：

与Single Number不同，这里不能再用异或操作来做了。可以为每个位上1的数目计数。然后将该计数%3，剩下的1肯定是单独的数字贡献的。

class Solution {public:    int singleNumber(vector<int>& nums) {        vector<int> bitNumCount(32, 0);        int len = nums.size();        for(int i=0; i<len; i++){            int num = nums[i];            for(int j = 0; j<32 && num!=0; j++){                if(num & 0x1 == 1){                    bitNumCount[j]++;                }                num = num>>1;            }        }        int result = 0;        for(int i=31; i>=0; i--){            result = result<<1;            if(bitNumCount[i]%3!=0){                result += 1;            }        }        return result;    }};

上面用了一个32位的向量。经过优化，还可以更为精简：

class Solution {public:    int singleNumber(vector<int>& nums) {        int result = 0;        int len = nums.size();        for(int i=0; i<32; i++){            int count = 0;            for(int j=0; j<len; j++){                count += (nums[j]>>i)&1;            }            result += (count%3)<<i;        }                return result;    }};

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