leetcode best time to buy and sell stock

1.一次买入一次卖出，O(n)算法

public class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length, minPrice = Integer.MAX_VALUE;
        int res =0 ;
        for(int i=0;i<n;i++){
            if(prices[i]<minPrice)
                minPrice = prices[i];
            else if(prices[i]-minPrice > res){
                res = prices[i]-minPrice;
            }
        }
        return res;
    }
}

2.多次买入，多次卖出，但是同时只能持有一个股票。此时只需要判断相邻两天是否递增即可

public class Solution {
    public int maxProfit(int[] prices) {
        int res = 0;
        for(int i=1;i<prices.length;i++){
            int tmp = prices[i]-prices[i-1];
            if(tmp>0)
                res += tmp;
        }
        return res;
    }
}

3.两次买入两次卖出，不能同时持有

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length==0) return 0;
        int n = prices.length;
        int buy1 = Integer.MIN_VALUE ,buy2 = Integer.MIN_VALUE , sell1 = 0, sell2 = 0;
        for(int i=0;i<n;i++){
            buy1 = Math.max(buy1,-prices[i]);  //加负号最大实质上是最小
            sell1 = Math.max(sell1,prices[i]+buy1);
            buy2 = Math.max(buy2,sell1-prices[i]);
            sell2 = Math.max(sell2,prices[i]+buy2);
        }
        return sell2;
    }
}