1. 162 Find Peak Element
   题意：返回数组中的其中一个peak
   思路：二分
2. 165 Compare Version Numbers
   题意：序列号比较
   思路：split后处理

class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector<string> vs1 = split_help(version1);
        vector<string> vs2 = split_help(version2);
        int index_1 = 0,size_1 = vs1.size();
        int index_2 = 0,size_2 = vs2.size();
        if(size_1 == 0) return -1;
        if(size_2 == 0) return 1;
        while(index_1 < size_1 && index_2 < size_2){
            if(vs1[index_1].size() == 0 && vs2[index_2].size() > 0){
                return -1;
            }else if(vs1[index_1].size() > 0 && vs2[index_2].size() == 0){
                return 1;
            }else if(vs1[index_1].size() > 0 && vs2[index_2].size() > 0){
                int tmp1 = 0,tmp2 = 0,i = 0,j = 0;
                while(i < vs1[index_1].size()){
                    tmp1 = tmp1 * 10 + vs1[index_1][i] - '0';
                    i ++;
                }
                while(j < vs2[index_2].size()){
                    tmp2 = tmp2 * 10 + vs2[index_2][j] - '0';
                    j ++;
                }
                if(tmp1 > tmp2) return 1;
                if(tmp1 < tmp2) return -1;
            }
            index_1 += 1;
            index_2 += 1;
        }
        while(index_1 < size_1){
            for(int i = 0;i < vs1[index_1].size();++ i) if(vs1[index_1][i] > '0') return 1;
            index_1 ++;
        }
        while(index_2 < size_2){
            for(int i = 0;i < vs2[index_2].size();++ i) if(vs2[index_2][i] > '0') return -1;
            index_2 ++;
        }
            
        // for(string s : vs2) cout<<s<<' ';
        return 0;
    }
    vector<string> split_help(string s){
        vector<string> ret;
        int index = 0,begin = 0;
        while(index < s.size()){
            while(s[index] != '.' && index < s.size()) index ++;
            // cout<<begin<<' '<<index<<endl;
            // cout<<s.substr(begin,index - begin)<<endl;
            ret.push_back(s.substr(begin,index - begin));
            index += 1;
            begin = index;
        }
        return ret;
    }
};

3. 166 Fraction to Recurring Decimal
   题意：除法模拟
   思路：map

class Solution
{
  public:
    string fractionToDecimal(long long numerator, long long denominator)
    {
        string result;
        stringstream ss;
        if ((numerator < 0 && denominator > 0) || (numerator > 0 && denominator < 0))
        {
            ss << '-';
        }
        numerator = abs(numerator);
        denominator = abs(denominator);
        long long intPart = numerator / denominator;
        long long r = numerator % denominator;
        ss << intPart;
        ss >> result;
        if (!r)
        {
            return result;
        }
        unordered_map<long long, size_t> table;
        result += '.';
        while (r)
        {
            if (table.find(r) != table.end())
            {
                size_t index = table.at(r);
                result.insert(index, 1, '(');
                result += ')';
                break;
            }
            else
            {
                table.emplace(r, result.size());
                char toAppend = r * 10 / denominator + '0';
                r = r * 10 % denominator;
                result += toAppend;
            }
        }
        return result;
    }
};

4. 167 Two Sum II - Input array is sorted
   题意：有序数组中找和为target的两个数
   思路：two point
5. 168 Excel Sheet Column Title
   题意：数字的变换
   这个傻逼题目真的是。。。

class Solution {
public:
    string convertToTitle(int n) {
        string ret = "";
        while(n){
            int curr = n % 26;
            // curr = curr == 0 ? 26 : curr;
            if(curr == 0){
                curr = 26;
                n -= 26;
            }
            ret += ('A' + curr - 1);
            n /= 26;
        }
        reverse(ret.begin(), ret.end());
        return ret;
    }

};

6. 169 Majority Element
   题意：出现概率大于0.5的数字