leetcode:122. Best Time to Buy and Sell Stock II-优快云博客

本文链接：https://blog.youkuaiyun.com/hunter___/article/details/89850779

博客围绕最佳买卖点问题展开，起初认为原内容复杂、条件不清，浪费时间，后提出用枚举法解决。还给出代码示例，并分析了不同序列下的利润计算，指出应关注单调序列。

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最佳买卖点：

写的啥，太复杂了

/*
thinking：
1.get the smaller num to buy, 
2.assume cur num to buy if next is smaller ,not buy,
3.when cur<before, cur<next buy,


================
for each num:
buy or not: not bought ,< pre,<next
sell or not:have bought, > pre,>next 

*/

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        auto sum=0,buy=0,sell=0;
        auto len=prices.size();
        auto cant_buy=false,cant_sell=false;
     
        if(len<2) return -1;

        for(auto i=0;i<len;i++)
        {
            // buy
            if(!cant_buy && i<len-1 && prices[i]<prices[i+1] )
            {            
                buy=prices[i];
                // sum-=buy;
                cant_buy=true;//can not buy
                cant_sell=false;//can sell
                printf("<=====buy:%d\n",buy);  
            }
            //sell
            if(!cant_sell && i<len-1 && prices[i]>prices[i+1] )
            {
             sell=prices[i];
                sum-=buy;
                sum+=sell;
                cant_sell=true;//can not sell
                cant_buy=false;//can buy
                printf("=>sell:%d\n",sell);
                
            }
            printf("====price:%d\n",prices[i]);
        }
           return sum; 
        
    }
};

,,,,写的不对，，

/*
thinking：
1.get the smaller num to buy, 
2.assume cur num to buy if next is smaller ,not buy,
3.when cur<before, cur<next buy,


================
for each num:
buy or not: not bought ,< pre,<next
sell or not:have bought, > pre,>next 

*/

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        auto sum=0,buy=0,sell=0;
        auto len=prices.size();
        auto have_buy=false,have_sell=true;
     
        if(len<2) return -1;

        buy=prices[0];
        for(auto i=0;i<len;i++)
        {
            if( buy>prices[i])
            {
                buy=prices[i]; 
                have_buy=true;
                // // sell=prices[i];
                // continue;
            }
            // have_buy=true;
         
            printf("==>buy:%d\n",buy);
            if( have_buy && sell< prices[i] )
            {     
                sell=prices[i];
                 printf("sell:%d\n",sell);
                // continue;
            }
             // have_sell=true; 
            
            if(sell_now)
            { 
                sum+=sell-buy;
                buy=prices[i+1];sell=prices[i+1];
            }
               printf("====price:%d\n",prices[i]);
        }
           return sum; 
        
    }
};

强迫症，浪费太多时间了。。。。

就是考虑不清楚，条件没写清。

枚举法呀。再来。

浪费时间：直接学习别人的吧。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        auto sum=0,buy=0,sell=0;
        auto len=prices.size();
        auto get_buy_p=false,get_sell_p=false;
     
        if(len<2) return -1;

        //find the reverse sort
        
        if(prices[0]<prices[1])
            buy=prices[0];
        
        buy=prices[0];
        for(auto i=0;i<len-1;i++)
        {         
            if( prices[i] < prices[i+1])
            { 
                buy=prices[i]; 
                get_buy_p=true;
            }
            
            
            if(get_buy_p && prices[i] > prices[i+1] )
            {     
                sell=prices[i];
                get_sell_p=true;
                 printf("sell:%d\n",sell);
            }

            
            if(get_sell_p && get_buy_p)
            { 
                sum+=sell-buy;
                get_sell_p=false;  get_buy_p=false;
                // buy=prices[i+1];sell=prices[i+1];
            }
               printf("====price:%d\n",prices[i]);
        }
           return sum; 
        
    }
};

看懂大神的再写

果然是技巧啊，，，，厉害：

First we post the code here.

int maxProfit(vector<int> &prices) {
    int ret = 0;
    for (size_t p = 1; p < prices.size(); ++p) 
      ret += max(prices[p] - prices[p - 1], 0);    
    return ret;
}

Second, suppose the first sequence is "a <= b <= c <= d", the profit is "d - a = (b - a) + (c - b) + (d - c)" without a doubt. And suppose another one is "a <= b >= b' <= c <= d", the profit is not difficult to be figured out as "(b - a) + (d - b')". So you just target at monotone sequences.