121. 买卖股票的最佳时机

简单题，很容易想到 O(n²)的代码：

 public int maxProfit(int[] prices) {
        int res = 0;

        for(int i = 0; i< prices.length;i++){
            for(int j = i+1; j<prices.length;j++){
                res = Math.max(res,prices[j] - prices[i]);
            }
        }
        return res;
    }

但是这样会超时，

public int maxProfit(int[] prices) {
        if(prices == null) return 0;
        int[] maxs = new int[prices.length];
        int res = 0;

        //记录前i-1天最小值
        int minLastI = prices[0];
        for(int i = 1; i< prices.length;i++){
            maxs[i] = Math.max(maxs[i-1],prices[i] - minLastI);
            if(prices[i] < minLastI){
                minLastI = prices[i];
            }
        }
        return maxs[prices.length-1];
    }

可以考虑继续优化空间

public int maxProfit(int prices[]) {
        int minprice = Integer.MAX_VALUE;
        int maxprofit = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minprice) {
                minprice = prices[i];
            } else if (prices[i] - minprice > maxprofit) {
                maxprofit = prices[i] - minprice;
            }
        }
        return maxprofit;
    }