LeetCode-137-Single Number II
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算法描述:
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
解题思路:用不进位加法模拟异或运算。这个题可以用累加和模3:由于数据要么出现三次,要么出现一次,所以,模3的结果只有两种,0或者1。由模后的结果重新构成的数字就是出现一次的数字。
int singleNumber(vector<int>& nums) { if(nums.size()== 1) return nums[0]; int res = 0; for(int i=0; i< 32; i++){ int sum = 0; for(int j=0; j < nums.size(); j++){ sum += (nums[j] >> i) & 1; sum %= 3; } res = res | (sum << i); } return res; }
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