题解 | #买卖股票的最好时机(四)#
买卖股票的最好时机(四)
https://www.nowcoder.com/practice/ba3c096c19e04afbbbd59250e909ac68
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class Solution {
public:
int solve(vector<int>& prices, int k) {
//dp[i][0],表示第i次买入操作收益
//dp[i][1],表示第i次卖出操作收益
vector<vector<int>> dp(k, vector<int>(2, 0));
for(int i = 0; i < k; ++i) {
dp[i][0] = -prices[0];
}
for (int i = 1; i < (int)prices.size(); ++i) {
for (int j = 0; j < k; ++j) {
if (j == 0) {
dp[j][0] = max(dp[j][0], -prices[i]); //在第i天进行第1次买入操作,最大收益
dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]);//在第i天进行第1次卖出操作,最大收益
} else {
dp[j][0] = max(dp[j][0], dp[j-1][1] - prices[i]);//在第i天进行第j次买入操作,最大收益
dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]); //在第i天进行第j次卖出操作,最大收益
}
}
}
return dp[k-1][1] > 0 ? dp[k-1][1] : 0;
}
};
int main() {
int n, k;
cin >> n >> k;
vector<int> prices(n, 0);
for (int i = 0; i < n; ++i)
cin >> prices[i];
Solution * s = new Solution();
int ret = s->solve(prices, k);
cout << ret << endl;
return 0;
}
#include<vector>
#include<algorithm>
using namespace std;
class Solution {
public:
int solve(vector<int>& prices, int k) {
//dp[i][0],表示第i次买入操作收益
//dp[i][1],表示第i次卖出操作收益
vector<vector<int>> dp(k, vector<int>(2, 0));
for(int i = 0; i < k; ++i) {
dp[i][0] = -prices[0];
}
for (int i = 1; i < (int)prices.size(); ++i) {
for (int j = 0; j < k; ++j) {
if (j == 0) {
dp[j][0] = max(dp[j][0], -prices[i]); //在第i天进行第1次买入操作,最大收益
dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]);//在第i天进行第1次卖出操作,最大收益
} else {
dp[j][0] = max(dp[j][0], dp[j-1][1] - prices[i]);//在第i天进行第j次买入操作,最大收益
dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]); //在第i天进行第j次卖出操作,最大收益
}
}
}
return dp[k-1][1] > 0 ? dp[k-1][1] : 0;
}
};
int main() {
int n, k;
cin >> n >> k;
vector<int> prices(n, 0);
for (int i = 0; i < n; ++i)
cin >> prices[i];
Solution * s = new Solution();
int ret = s->solve(prices, k);
cout << ret << endl;
return 0;
}