Leetcode-single number II

Leetcode-single number II

Given an array of integers, every element appears three times except for one. Find the single one.

Note:
Your algorithm should has a linear runtime complexity. Could you implement it without using extra memory?

classSolution { Public:    intSinglenumber (vector<int>&nums) {        intres =0; intn[ +]={0};  for(inti =0; I < nums.size (); i++){           for(intj =0; J < +; J + +) {N[j]+ = (nums[i]>>j) &1; }        }         for(intj =0; J < +; J + +) {res|= (n[j]%3) <<J; }        returnRes; }};

Use a 32-dimensional array to store the number of occurrences on each of the 1, and finally the number of each of the 3 to the remainder, for the number to be the number on that,

For example, the 5th place gets 21, except 3 to take the remainder 0, indicating that the number to be asked in the 5th place is 0

Finally, the number can be recovered.

Complexity of Time: O (32*n)

http://blog.csdn.net/jiadebin890724/article/details/23306837

Leetcode-single number II