Leetcode 188 买卖股票的最佳时机 Ⅳ

Wils0nEdwards

于 2025-03-26 15:35:24 发布

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文章标签： leetcode 算法职场和发展

本文链接：https://blog.youkuaiyun.com/coldasice342/article/details/146531849

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java solution

class Solution {
   
    public int maxProfit(int k, int[] prices) {
   
        int n = prices.length;
        if(n == 1) return 0;
        if(k >= n/2) {
   
            return maxProfitUnlimited(prices);
        }
        //创建dp数组
        int[][] dp = new int[k + 1][n];

        for(int i = 1; i <= k; i++) {
   
            int maxDiff = -prices[0]; //最开始dp[0][j] = 0, 所以 maxDiff 初始值为 -prices[0]
            for(int j = 1; j < n; j++) {
   
                dp[i][j] = Math.max(dp[i][j - 1], prices[j] + maxDiff);
                maxDiff = Math.max(maxDiff, dp[i - 1][j] - prices[j]);
//dp[i][j] = Math.max(dp[i][j - 1], \sigma 1-m (prices[j] - prices[m] + dp[i - 1][m]))
//本来dp[i][j]是这个状态转移方程，为了优化计算，我们把prices[j]提出来,剩余部分定义为maxDiff
            }
        }

        return dp[k][n - 1];
    }

    private int maxProfitUnlimited(int[] prices) {
   
        int profit = 0;
        for(int i = 1; i < prices.length; i++) {
   
            if(prices[i] > prices[i - 1]) {
   
                profit