股票盈利问题

笔试遇到好几次。

方法一：类似于动态规划？暴力法，递归层级计算最优解

public int calculate(int prices[], int s) {
        if (s >= prices.length)
            return 0;
        int max = 0;
        for (int start = s; start < prices.length; start++) {
            int maxprofit = 0;
            for (int i = start + 1; i < prices.length; i++) {
                if (prices[start] < prices[i]) {
                    int profit = calculate(prices, i + 1) + prices[i] - prices[start];
                    if (profit > maxprofit)
                        maxprofit = profit;
                }
            }
            if (maxprofit > max)
                max = maxprofit;
        }
        return max;
    }
}

方法二：数学分析

 寻找递增的序列，并让峰值和谷值进行相减。

class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length==0||prices==null)
    		return 0;
        return cal(prices, 0);
    }
    public int cal(int prices[],int s){
    	int len=prices.length;
    	int j=0;
    	int count=0;
    	for(int i=1;i=prices[i-1]){
    			continue;
    		}
    		count+=(prices[i-1]-prices[j]);
    		j=i;
    	}
    	if(j!=len-1){
    		count+=(prices[len-1]-prices[j]);
    	}
    	return count;
    }
}