Best Time to Buy and Sell Stock III

最新推荐文章于 2022-04-26 10:52:27 发布

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本文探讨了在股票市场中通过最多两次交易获取最大利润的算法。提出了一种改进的时间复杂度为O(n)的方法，通过动态规划简化了问题解决过程，避免了重复计算，有效提高了效率。

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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 two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思路一，既然是两次交易，那么我就求所有可能的两次交易，即profit=max{(0,i),(i,n)},0<=i<n,最终时间复杂度为O(n2)

<span style="font-size:18px;">public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length == 0){  
            return 0;  
        } 
        int profit = 0;
        for(int i = 0; i < prices.length-1 ; i ++){
            profit = pro(0,i,prices) + pro(i+1,prices.length-1,prices) > profit ? pro(0,i,prices) + pro(i+1,prices.length-1,prices) : profit;
        }
        return profit;
    }
    public int pro(int i, int j,int[] prices){//i到j这几天的最大收益
        if(i == j){
            return 0;
        }
        int buy =  prices[i];
        int sell = prices[j];
        int profit = sell - buy;
        int index = i + 1;
        while(index <= j){
            if(prices[index] > sell){
                sell = prices[index];
                profit = sell - buy > profit? sell - buy: profit;
            }else if(prices[index] < buy){
                sell = prices[index];
                buy = prices[index];
            }
            index ++;
        }
        return profit;
    }
}</span>

思路二：

感觉是DP，但是怎么也想不到，试了试解法一，结果不出所料，Time Limit Exceeded

考虑解法一的缺点，当求第i天的收益时，直接就是按不就班的从第0天计算到第i天，那么计算所得的第i-1天的收益直接丢弃不使用，那么思考一下，是否可以利用辛辛苦苦得到的第i-1天收益的数据？答案是可以，如果我们记录了第i-1天之前的最小值min，那么第i天的收益该怎么计算，一定是DP[i]=max(DP[i-1],prices[i]-min)。

http://blog.unieagle.net/2012/12/05/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Abest-time-to-buy-and-sell-stock-iii%EF%BC%8C%E4%B8%80%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length <= 1){  
            return 0;  
        } 
        int[] left = new int[prices.length];//存储从左向右扫描得到的第i天最大的收益
        left[0] = 0;
        int min = prices[0];//第i天之前最小的值
        //需找DP方程
        //当prices[i] - min > left[i-1] 时 left[i] = prices[i] - min;
        //当prices[i] - min < left[i-1] 时 left[i] = left[i-1],且min = MIN(prices[i],min)
        for(int i = 1; i < prices.length ; i ++){
            left[i] = prices[i] - min > left[i-1] ? prices[i] - min : left[i-1];
            min = prices[i] < min ? prices[i]: min;
        }
        //同理
        int[] right = new int[prices.length];//存储从右向左扫描得到的第i天的最大收益
        right[prices.length-1] = 0;
        int max = prices[prices.length - 1];
        for(int i = prices.length - 2 ; i >= 0 ; i--){
            right[i] = max - prices[i] > right[i+1] ? max - prices[i] : right[i+1];
            max = prices[i] > max ? prices[i] : max;
        }
        int profit = left[0] + right[0];
        for(int i = 1 ; i < prices.length ; i ++){
            if(left[i] + right[i] > profit){
                profit = left[i] + right[i];
            }
        }
        return profit;
    }
}