137. Single Number II
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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?
Analysis:
参考:http://www.acmerblog.com/leetcode-single-number-ii-5394.html
只写了一个中等方法,但更容易理解,普适性更强些。
Source Code(C++):
#include <iostream>#include <vector>using namespace std;class Solution {public: int singleNumber(vector<int>& nums) { if (nums.size()<1) {// return ; } int bit_nums=0, res=0; for(int i=0; i<32; i++) { for (int j=0; j<nums.size(); j++) { if (((nums.at(j)>>i)&1) == 1) { bit_nums++; } } res |= (bit_nums%3)<<i; bit_nums=0; } return res; }};int main() { vector<int> v; v.push_back(5); v.push_back(5); v.push_back(5); v.push_back(10000); Solution sol; cout << sol.singleNumber(v); return 0;}
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- 137.Single Number II
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- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
- 137. Single Number II
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