version at: 2022-04-21

/**
1. 找到每一段连续上升子数组的差值
2. 尝试从某个数开始向右扩展, 直到下一个数 a[i+1] < a[i] 时停止, 并记录
3. 结果为所有差值的和
*/

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null) return 0;

        int sum = 0;

        for (int i = 0, j = 0; j < prices.length; j++) {
            if (j == prices.length-1 || prices[j] > prices[j+1]) {
                sum += prices[j] - prices[i];                
                i = j+1;
            }
        }

        return sum;
    }
}

/*
1. 只能串行的买卖 -> 找出所有连续上升区间并累加所有 区间尾减区间头的差
*/

class Solution {
    public int maxProfit(int[] prices) {
        int i = 0,j = 0, n = prices.length;
        int res = 0;

        while (j < n && i < n){
            if (j == n-1||  prices[j] > prices[j+1] ){
                res += prices[j] - prices[i];
                i = j+1;
            }
            j++;
        }
        return res;
    }
}