Skip to content

Latest commit

 

History

History
124 lines (92 loc) · 4.05 KB

0188-best-time-to-buy-and-sell-stock-iv.adoc

File metadata and controls

124 lines (92 loc) · 4.05 KB

188. Best Time to Buy and Sell Stock IV

{leetcode}/problems/best-time-to-buy-and-sell-stock-iv/[LeetCode - Best Time to Buy and Sell Stock IV^]

Say you have an array for which the ith 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.
{image_attr} 
public int maxProfit(int maxK, int[] prices) {
    if (Objects.isNull(prices) || prices.length == 0 || maxK == 0) {
        return 0;
    }
    if (maxK > prices.length / 2) { // (1)
        return maxProfit(prices);
    }
    int[][][] dp = new int[prices.length][maxK + 1][2]; // (2)
    for (int i = 0; i < prices.length; i++) {
        for (int k = maxK; k >= 1; k--) {
            if (i == 0) {
                dp[i][k][0] = 0;
                dp[i][k][1] = -prices[i]; // (3)
                continue;
            }
            dp[i][k][0] = Math.max(dp[i - 1][k][0], dp[i - 1][k][1] + prices[i]);
            dp[i][k][1] = Math.max(dp[i - 1][k][1], dp[i - 1][k - 1][0] - prices[i]);
        }
    }
    return dp[prices.length - 1][maxK][0];
}

public int maxProfit(int[] prices) {
    int dp0 = 0;
    int dp1 = Integer.MIN_VALUE;
    for (int i = 0; i < prices.length; i++) {
        int temp = dp0; // (4)
        dp0 = Math.max(dp0, dp1 + prices[i]);
        dp1 = Math.max(dp1, temp - prices[i]);  // (4)
    }
    return dp0;
}

  1. 为什么只需要大于 prices.length / 2 时就可以调用没有限制的函数？

  2. 为什么要把数组大小设置为 maxK + 1，而不是 maxK？

  3. 为什么有时初始化成 -prices[i]，而有时又是 Integer.MIN_VALUE？

  4. 为什么有时需要保存这个变量？而有时又不需要？

思考题：再仔细思考思考这个解题框架。

参考资料

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.
link:{sourcedir}/_0188_BestTimeToBuyAndSellStockIV.java[role=include]