问题描述
解题思路
123.买卖股票的最佳时机III相当于是本题中\(k=2\)的情形,递推关系的推导与写法参照123.买卖股票的最佳时机III即可。
代码
#include <vector>
using std::vector;
class Solution {
public:
int maxProfit(int k, vector<int> &prices) {
vector<vector<int>> dp(prices.size() + 1, vector<int>(2 * k + 1, 0));
for (int j = 1; j <= 2 * k; j++) {
if (j % 2 == 1)
dp[0][j] = -prices[0];
}
for (int i = 1; i <= prices.size(); i++) {
for (int j = 1; j <= 2 * k; j++) {
if (j % 2 == 1)
dp[i][j] = max(dp[i - 1][j - 1] - prices[i - 1], dp[i - 1][j]);
else
dp[i][j] = max(dp[i - 1][j - 1] + prices[i - 1], dp[i - 1][j]);
}
}
return dp[prices.size()][2 * k];
}
};
标签:sell,buy,买卖,int,vector,iv,最佳时机,188,IV
From: https://www.cnblogs.com/zwyyy456/p/16792267.html