2015-09-09 [一点资讯]--数据抓取和处理工程师--4面

时间：2015-09-09 14:40 ~ 15:40

地点：北京市海淀区王庄路1号 清华同方科技广场D座 西区 7层

1. 问项目经验和看过什么书

2. 有3N+1个整数，其中只有一个数出现了一次，其它的数都出现了3次，查找唯一出现一次的那个数。

https://leetcode.com/problems/single-number-ii/

https://github.com/loverszhaokai/leetcode/blob/master/137_no/sol_bit_manipulation.cc

#include <climits>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <iomanip>
#include <string>
#include <vector>

using namespace std;

int find(const vector<int> &nums)
{
    int count[32];

    for (int iii = 0; iii < sizeof(count) / sizeof(int); iii++)
        count[iii] = 0;

    for (int iii = 0; iii < nums.size(); iii++) {

        int jjj = 1;
        int kkk = 1;

        while (jjj <= 32) {

            if (nums[iii] & kkk)
                count[jjj - 1] = (count[jjj - 1] + 1) % 3;

            kkk <<= 1;
            jjj++;
        }
    }

    int ans = 0;
    int jjj = 1;

    for (int iii = 0; iii < sizeof(count) / sizeof(int); iii++) {

        if (count[iii])
            ans |= jjj;

        jjj <<= 1;
    }

    return ans;
}

// Use two integer instead of count[32]
// 0 0 -> 0
// 0 1 -> 1
// 1 0 -> 2
//
int find2(const vector<int> &nums)
{
    int high = 0;
    int low = 0;

    for (int iii = 0; iii < nums.size(); iii++) {

        int jjj = 1;
        int kkk = 1;

        while (jjj <= 32) {

            if (nums[iii] & kkk) {

                if ((high & kkk) ^ (low & kkk) ^ kkk)
                    low |= kkk;
                else
                    low &= ~kkk;

                if ((high & kkk) ^ (low & kkk) ^ kkk)
                    high |= kkk;
                else
                    high &= ~kkk;
            }

            kkk <<= 1;
            jjj++;
        }
    }

    int ans = 0;
    int iii = 1;
    int jjj = 1;

    while (iii <= 32) {

        if (low & jjj || high & jjj)
            ans |= jjj;

        jjj <<= 1;
        iii++;
    }

    return ans;
}


int main()
{
    const struct TestCase {
        vector<int> nums;
        int ret;
    } test_cases[] = {
        {
            { 1, 1, 1, 3 },
            3
        },
        {
            { 1, 1, 1, 2, 3, 2, 2 },
            3
        },
        {
            { 1, 1, 1, 2, 3, 2, 2, 3, 4, 3 },
            4
        },
    };

    for (int iii = 0; iii < sizeof(test_cases) / sizeof(TestCase); iii++) {

        const TestCase &tc = test_cases[iii];

        int actual_ret = find(tc.nums);

        if (tc.ret != actual_ret) {
            cout << "Case #" << iii << ": FAILED" <<  endl;
            cout << "	Expected ret=" << tc.ret << endl;
            cout << "	Acutal   ret=" << actual_ret << endl;
        }

        actual_ret = find2(tc.nums);

        if (tc.ret != actual_ret) {
            cout << "Case #" << iii << ": FAILED" <<  endl;
            cout << "	Expected ret=" << tc.ret << endl;
            cout << "	Acutal   ret=" << actual_ret << endl;
        }

    }

    return 0;
}

3. 问github上的项目，推荐TypoChecker使用Trie树。