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题目描述

Say you have an array for which the i-th element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Example 1:
Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

代码

Approach #1 Dynamic Programming

public class Solution {
public int maxProfit(int k, int[] prices) {
if (prices == null || prices.length <= 1 || k <= 0) return 0;
// 1. Special case
int n = prices.length;
if (k >= n / 2) {
int profit_max = 0;
for (int i = 1; i < n; i++) {
if (prices[i] - prices[i - 1] > 0) {
profit_max += prices[i] - prices[i - 1];
}
}
return profit_max;
}
// 2. dp
int[][] f = new int[n + 1][k + 1];
for (int j = 1; j <= k; j++) {
for (int i = 1; i <= n; i++) {
for (int x = 0; x <= i; x++) {
f[i][j] = Math.max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i));
}
}
}
return f[n][k];
}
// [left, right]一次交易最大收益
private int profit(int[] prices, int left, int right) {
if (left >= right) return 0;
int leftMin = Integer.MAX_VALUE;
int profit = 0;
for (int i = left; i < right; i++) {
profit = Math.max(profit, prices[i] - leftMin);
leftMin = Math.min(leftMin, prices[i]);
}
return profit;
}
}