给定一个整数数组 prices ,它的第 i 个元素 prices[i] 是一支给定的股票在第 i 天的价格。
设计一个算法来计算你所能获取的最大利润。你最多可以完成 k 笔交易。
注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。
class Solution {
private static int all(int[] prices) {
int ret = 0;
for (int i = 1; i < prices.length; ++i) {
ret += Math.max(0, prices[i] - prices[i - 1]);
}
return ret;
}
public static int maxProfit0(int k, int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int n = prices.length;
int[][] dp = new int[n][k + 1];
/**
* dp[i][j] = dp[i - 1][j]
*
* dp[i][j - 1] + prices[i] - prices[i]
* dp[i - 1][j - 1] + prices[i] - prices[i - 1]
* dp[i - 2][j - 1] + prices[i] - prices[i - 2]
* dp[i - 3][j - 1] + prices[i] - prices[i - 3]
*/
for (int i = 1; i < n; ++i) {
for (int j = 1; j <= k; ++j) {
int max = 0;
for (int s = 0; s <= i; ++s) {
max = Math.max(max, dp[s][j - 1] + prices[i] - prices[s]);
}
dp[i][j] = Math.max(dp[i - 1][j], max);
}
}
return dp[n - 1][k];
}
public static int maxProfit(int k, int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int n = prices.length;
if (k >= n / 2) {
return all(prices);
}
int[][] dp = new int[n][k + 1];
/**
* dp[i][j] = dp[i - 1][j]
*
* dp[i][j - 1] + prices[i] - prices[i]
* dp[i - 1][j - 1] + prices[i] - prices[i - 1]
* dp[i - 2][j - 1] + prices[i] - prices[i - 2]
* dp[i - 3][j - 1] + prices[i] - prices[i - 3]
*/
for (int j = 1; j <= k; ++j) {
int pre = dp[0][j - 1] - prices[0];
for (int i = 1; i < n; ++i) {
dp[i][j] = Math.max(dp[i - 1][j], Math.max(pre, dp[i][j - 1] - prices[i]) + prices[i]);
pre = Math.max(pre, dp[i][j - 1] - prices[i]);
}
}
return dp[n - 1][k];
}
public static void main(String[] args) {
System.out.println(maxProfit(2, new int[]{2, 4, 1}));
}
}