LeetCode（算法）- 121. 买卖股票的最佳时机

// 解决方案(1)
class Solution {
    public int maxProfit(int[] prices) {
        int cost = Integer.MAX_VALUE, profit = 0;
        for(int price : prices) {
            cost = Math.min(cost, price);
            profit = Math.max(profit, price - cost);
        }
        return profit;
    }
}
// 解决方案(2)
class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 0) {
            return 0;
        }
        List<Integer> list = new ArrayList<>();
        // 当前买入最划算指针
        int ptr = 0;
        list.add(prices[ptr]);
        int maxn = prices[ptr];
        for (int i = 1; i < prices.length; i++) {
            int num = prices[i];
            // 当天卖出大于之前的最大卖出值, 直接找到买入最划算指针
            if (num > maxn) {
                maxn = num;
                while (ptr + 1 < list.size()) {
                    ptr++;
                }
            }
            // 当天卖出小于等于之前的最大卖出值, 遍历找出是否存在买入最划算候选指针, 比当前还要划算的
            else {
                for (int j = ptr; j < list.size(); j++) {
                    if (num - list.get(j) > maxn - list.get(ptr)) {
                        maxn = num;
                        ptr = j;
                    }
                }
            }
            // 如果比集合最小的还要小, 说明符合买入最划算的候选指针, 使用 LinkedList 则超时, 获取操作你懂的
            if (num < list.get(list.size() - 1)) {
                list.add(num);
            }
        }
        return maxn - list.get(ptr);
    }
}

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int cost = INT_MAX, profit = 0;
        for(int price : prices) {
            cost = min(cost, price);
            profit = max(profit, price - cost);
        }
        return profit;
    }
};