On到O1的方法

1. 单调栈（队列）

Leetcode 题目----含特定字母的最小子序列

class Solution {
    struct node{
        int id;
        char ch;
        node(int a,char c):id(a),ch(c){}
    };
    bool compare(node n1,node n2){
        return n1.ch > n2.ch;
    }
public:
    string smallestSubsequence(string s, int k, char letter, int repetition) {
        int n = s.size();
        vector<int>v(1,0);   //表示第i个letter在s中出现的位置
        int appear_num = 0;      //表示letter出现的次数
        for(int i = 0;i<n;i++){
            if(s[i] == letter){
                appear_num++;
                v.push_back(i);
            }
        }
        int other_character_num = k-repetition;
        string ret = "";
        int start_id = 0;
        int end_id = n-1;
        deque<node>dq;//相当于一个递增栈，每次从栈底贪心选出当前未知的字母。
        for(int i = 1;i<=other_character_num;i++){
            //总共有appear_num个letter,那么第1个lette出现的范围应该保证
            //1.选了之后还剩repetition-1个letter，如果目前的letter数量不够的话，够了就不管他了
            //2.选了之后至少还有k-1个字母
            int tmp1 = n-k;
            int tmp2 = func(n,appear_num,repetition,v);
            //cout<<"tmp1: "<<tmp1<<"tmp2: "<<tmp2<<endl;
            end_id = min(tmp1,tmp2);
            for(int j = start_id;j<=end_id;j++){
                if(dq.empty()){
                    dq.push_back(node(j,s[j]));
                    continue;
                }
                node n2 = node(j,s[j]);
                while(!dq.empty()){
                    node n1 = dq.back();
                    if(compare(n1,n2)){
                        dq.pop_back();
                    }else{
                        break;
                    }
                }
                dq.push_back(n2);
            }
            node target = dq.front();
            dq.pop_front();
            char ch = target.ch;
            ret += ch;
            start_id = end_id+1;
            //cout<<"ch:"<<ch<<" start_id:"<<start_id<<" size:"<<dq.size()<<endl;
            k--;
            if(ch == letter && repetition>0){
                repetition--;
                i--;
            }
        }
        while(repetition--){
            ret += letter;
        }
        return ret;
    }
private:
    int func(int n,int appear_num,int repetition,vector<int>& v){
        if(repetition<=0){
            return n-1;
        }else{
            return v[appear_num-repetition+1];
        }
    }   
};

2. 哈希表加数学推导（一般适用于等式的数学推导）

3. 动态规划

3.1类型 ：柱状图
接雨水1
柱状图中最大矩形，本题也可以用单调栈，需要分析一下。

4.bitset的使用

大佬题解

5.双指针

适合于求连续区间的长度的问题，即窗口大小问题。
考试最大困扰度