leetcode轮回计划20181128

1. 382 Linked List Random Node
   题意：在不知道长度的链表中随机采样
   思路：蓄水池采样
2. 384 Shuffle an Array
   题意：洗牌
   思路：注意是i + rand() % (res.size() - i)而不是rand() % res.size()。即：从拿到的牌中选牌来交换而非整副牌中选牌来交换。
3. 385 Mini Parser
   思路：关于stringsteam的简单使用

class Solution {
public:
    NestedInteger deserialize(string s) {
        istringstream in(s);
        return deserialize(in);
    }
private:
    NestedInteger deserialize(istringstream &in){
        int num;
        if(in >> num)
            return NestedInteger(num);
        in.clear();
        in.get();
        NestedInteger list;
        while(in.peek() != ']'){
            list.add(deserialize(in));
            if(in.peek() == ',')
                in.get();
        }
        in.get();
        return list;
    }
};

4. 386 Lexicographical Numbers
   题意：字典顺序返回1-n
   思路：找规律
5. 388 Longest Absolute File Path
   题意：最长文件路径长度
   思路：stringsteam的简单使用

class Solution {
public:
    int lengthLongestPath(string input) {
        vector lines;
        string s_tmp;
        stringstream ss(input);
        while(getline(ss, s_tmp, '\n')) lines.push_back(s_tmp);
        map tool;
        int maxlen = 0;
        for(auto line : lines){
            int last = line.find_first_not_of('\t');
            int length = line.size() - last;
            if(line.find('.') != -1){
                maxlen = max(maxlen, tool[last] + length);
            }else{
                tool[last + 1] = tool[last] + length + 1;
            }
        }
        return maxlen;
    }
};

6. 390 Elimination Game
   题意：关灯实验，顺序为左右和右左交替。返回最后剩下的等的序号
   思路：分治。注意右左的时候需要考虑奇偶的问题

class Solution {
public:
    int lastRemaining(int n) {
        return help(n, true);
    }
    int help(int n, bool left2right){
        return n == 1 ? 1 : left2right ? 2 * help(n / 2, false)
            : 2 * help(n / 2, true) - 1 + (n & 1);
    }
};