标签：Sell Buy 0123 int 复杂度 th prices day dp
1. 题目描述
2. Solution 1
1、思路分析
DP
1> 状态定义: dp[k][i] 表示在第i天进行第k笔交易能获得的最大利润。
2> 边界: dp[k][i] = {0}
3> 状态转移:
若在第i天没有进行交易，则最大利润为前一天的最大利润。即dp[k][i] = dp[k][i-1]。
若在第j天买入了股票，j = [0, ..., i-1]，然后在第i天卖出，利润为: prices[i] - prices[j] + dp[k-1][j-1]。
实际上，j也可以取到i，此时，利润为prices[i] - prices[j] + dp[k-1][i]。
综上: dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j-1]), j=[0..i-1]
2、代码实现
package Q0199.Q0123BestTimetoBuyandSellStockIII;
/*
It's not difficult to get the DP recursive formula:
dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j-1]), j=[0..i-1]
For k transactions, on i-th day,
if we don't trade then the profit is same as previous day dp[k, i-1];
and if we bought the share on j-th day where j=[0..i-1], then sell the share on i-th day then the profit is
prices[i] - prices[j] + dp[k-1, j-1] .
Actually j can be i as well. When j is i, the one more extra item
prices[i] - prices[j] + dp[k-1, j] = dp[k-1, i] looks like we just lose one chance of transaction.
I see someone else use the formula dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j]), where the
last one is dp[k-1, j] instead of dp[k-1, j-1]. It's not the direct sense, as if the share was bought on j-th
day, then the total profit of previous transactions should be done on (j-1)th day. However, the result based on
that formula is also correct, because if the share was sold on j-th day and then bought again, it is the same if
we didn't trade on that day.
So the straightforward implementation is:
*/
public class Solution1 {
public int maxProfit(int[] prices) {
if (prices.length == 0) return 0;
int[][] dp = new int[3][prices.length];
for (int k = 1; k <= 2; k++) {
for (int i = 1; i < prices.length; i++) {
int min = prices[0];
for (int j = 1; j <= i; j++)
min = Math.min(min, prices[j] - dp[k - 1][j - 1]);
dp[k][i] = Math.max(dp[k][i - 1], prices[i] - min);
}
}
return dp[2][prices.length - 1];
}
// Time complexity is O(kn^2), space complexity is O(kn).
}
3、复杂度分析
时间复杂度: O(kn^2)
空间复杂度: O(kn)
3. Solution 2
1、思路分析
Solution 1是进行k次交易的通用方法，本题中k=2，所以把数组拆成变量，可以降低空间复杂度。
2、代码实现
package Q0199.Q0123BestTimetoBuyandSellStockIII;
// In this case, K is 2. We can expand the array to all named variables:
public class Solution4 {
public int maxProfit(int[] prices) {
int buy1 = Integer.MAX_VALUE, buy2 = Integer.MAX_VALUE;
int sell1 = 0, sell2 = 0;
for (int price : prices) {
buy1 = Math.min(buy1, price);
sell1 = Math.max(sell1, price - buy1);
buy2 = Math.min(buy2, price - sell1);
sell2 = Math.max(sell2, price - buy2);
}
return sell2;
}
}
3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(1)
标签：Sell,Buy,0123,int,复杂度,th,prices,day,dp 来源： https://www.cnblogs.com/junstat/p/16283226.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。