137. Single Number II 难度:medium
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题目:
Given an array of integers, every element appears three times except for one. Find that single one.
思路:
那么对这32位中的每一位做相同的处理,也就是说,逐位把所有的输入加起来,并且看看第i位的和除以3的余数,这个余数就是single numer在第i位的取值。这样就得到了single number在第i位的取值。
程序:
class Solution {public: int singleNumber(vector<int>& nums) { int count[32]={0}; int result=0; int n = nums.size(); for(int i=0;i<32;i++){ for(int j=0;j<n;j++){ count[i]+=((nums[j]>>i)&1); count[i]=count[i]%3; } result|=(count[i]<<i); } return result; }};
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