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Link to the corresponding question:188. The best time to buy and sell stocks IV - LeetCode

Niuke corresponding question link:The best time to buy and sell stocks (Part 4)_Niuketiba_Niuke.com (nowcoder.com)

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1. Analyze the topic

1、Status indication

To make a clearer distinctionBuy it inandSell, we replaceHave stocksandNo stockTwo states.

• f[i][j] means: i After the day, it's done j transactions, at this timeHave stocksMaximum benefit of the state.
• g[i][j] means: i After the day, it's done j transactions, at this timeNo stockMaximum benefit of the state.

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2. State transfer equation

for f[i][j]There are also two situations in which i After the day is over, complete j The transaction is now in handHave stocksstatus:

• exist i-1 Day, handHave stocks, and traded j On the i-th day, do nothing.The profit at this time is f[i - 1][j]。
• exist i-1 Day, handNo stocks, and traded j times. i I bought shares on that day.Tickets. Then i At the end of the day, there are stocks. The profit at this time is g[i - 1][j] - prices[i]。

In the above two cases, what we need isMaximum, so the state transfer equation of f is:

f[i][j] = max(f[i - 1][j], g[i - 1][j] - prices[i]

for g[i][j], we have the following two situations in the first i After the day is over, complete j The transaction is now in handNo stocksstatus:

• exist i-1 Days, no stocks in hand, and traded j On the i-th day, do nothing.The profit at this time is g[i - 1][j]。
• exist i-1 Days ago, I had stocks in my hands and traded them. j - 1 times. i When the day comes,The stock is sold. i At the end of the day, we traded. j times. The profit at this time is f[i - 1][j - 1] + prices[i]。

In the above two cases, what we need isMaximum,therefore g The state transfer equation is:

g[i][j] = max(g[i - 1][j], f[i - 1][j - 1] + prices[i])

The transaction relationship between them is as follows:

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3. Initialization

• When in the 0 Days, only inBought onceThe state, the profit at this time is -prices[0] ,becausethis f[0][0] = - prices[0] 。
• In order to obtain max When some non-existent statesNo interferenceWe initialize them all to -INF(useINT_MIN In the calculation process, there will beoverflowThe risk here INF Half price 0x3f3f3f3f , small enough).

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4、Filling order

Fill in each row from top to bottom, and each row from left to right, filling in both tables together.

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5、return value

Returns the maximum value in the sell state, but we don't know how many times it was traded, so we return g The maximum value in the last row of the table.

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6、optimization

Our number of transactions will not exceed half of the total number of days, so we can first k Let’s deal with it and optimize the scale of the problem:k = min(k, n / 2)。

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2. Code

//力扣
//【动态规划-二维dp-2个状态】
class Solution {
    //f[i][j]:第i天结束后，完成了j次交易，此时处于“买入”状态下的最大利润
    //g[i][j]:第i天结束后，完成了j次交易，此时处于“卖出”状态下的最大利润
private:
    const int INF=0x3f3f3f3f;
public:
    int maxProfit(int k, vector<int>& prices) {
        int n=prices.size();
        k=min(k, n/2); //优化:处理最多交易次数
        vector<vector<int>> f(n, vector<int>(k+1, -INF));
        vector<vector<int>> g(n, vector<int>(k+1, -INF));
        f[0][0]=-prices[0], g[0][0]=0;
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<=k; j++)
            {
                f[i][j]=max(f[i-1][j], g[i-1][j]-prices[i]);
                g[i][j]=g[i-1][j];
                if(j>=1) g[i][j]=max(g[i][j], f[i-1][j-1]+prices[i]);
            }
        }
        int res=0;
        for(int j=0; j<=k; j++)
            res=max(res, g[n-1][j]);
        return res;
    }
};
 
//【动态规划-二维dp-2k+1个状态】
class Solution {
    //dp[i][0] -- 没有操作
    //下面j为奇数：买入；j为偶数：卖出 (j的范围：1~2k-1)
    //dp[i][j] -- 第1~k次买入
    //dp[i][j+1] -- 第1~k次卖出
public:
    int maxProfit(int k, vector<int>& prices) {
        int n=prices.size();
        vector<vector<int>> dp(n, vector<int>(2*k+1));
        for(int j=1; j<2*k; j+=2)
            dp[0][j]=-prices[0];
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<2*k; j+=2)
            {
                dp[i][j+1]=max(dp[i-1][j+1], dp[i-1][j]-prices[i]);
                dp[i][j+2]=max(dp[i-1][j+2], dp[i-1][j+1]+prices[i]);
            }
        }
        return dp[n-1][2*k];
    }
};

//牛客
#include <iostream>
#include <cstring>
using namespace std;
 
const int INF=0x3f3f3f3f;
const int N=1010, M=110;
int prices[N];
int f[N][M], g[N][M];
//f[i][j]:第i天结束后，完成了j次交易，此时处于“买入”状态下的最大利润
//g[i][j]:第i天结束后，完成了j次交易，此时处于“卖出”状态下的最大利润
 
int main()
{
    int n, k;
    cin >> n >> k;
    for(int i=0; i<n; i++)
        cin >> prices[i];
    memset(f, -INF, sizeof(f));
    memset(g, -INF, sizeof(g));
    int res=0;
    f[0][0]=-prices[0], g[0][0]=0;
    for(int i=1; i<n; i++)
    {
        for(int j=0; j<=k; j++)
        {
            f[i][j]=max(f[i-1][j], g[i-1][j]-prices[i]);
            g[i][j]=g[i-1][j];
            if(j>0) g[i][j]=max(g[i][j], f[i-1][j-1]+prices[i]);
            res=max(res, g[i][j]);
        }
    }
    cout << res << endl;
    return 0;
}
 
//值得学习的代码
#include <iostream>
using namespace std;
 
const int N = 1010, M = 110;
 
int n, k, p[N];
int f[N][M], g[N][M];
 
int main()
{
    cin >> n >> k;
    for(int i = 0; i < n; i++) cin >> p[i];
 
    k = min(k, n / 2);
    for(int j = 0; j <= k; j++) f[0][j] = g[0][j] = -0x3f3f3f3f;
    f[0][0] = -p[0], g[0][0] = 0;
 
    for(int i = 1; i < n; i++)
    {
        for(int j = 0; j <= k; j++)
        {
            f[i][j] = max(f[i - 1][j], g[i - 1][j] - p[i]);
            g[i][j] = g[i - 1][j];
            if(j >= 1) g[i][j] = max(g[i][j], f[i - 1][j - 1] + p[i]);
        }
    }
 
    int ret = 0;
    for(int j = 0; j <= k; j++) ret = max(ret, g[n - 1][j]);
 
    cout << ret << endl;
 
    return 0;
}

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3. Reflection and Improvement

For a series of questions like buying and selling stocks, you need to thoroughly master the core knowledge points.