LeetCode122. Best Time to Buy and Sell Stock II

122.Best Time to Buy and Sell Stock II

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

方法一:
这里与LeetCode121. Best Time to Buy and Sell Stock不同的是：可以多次买卖股票，但是每次买之前要把之前买的卖出去，这样的限定条件就让问题简化了。只要后面的价格比前面高，就会赚差价。

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

复杂度分析：
时间复杂度：O(n)
空间复杂度：O(1)

方法二：
(Peak Valley Approach) [Accepted]

public int maxProfit3(int[] prices) {    int i = 0;    int valley = prices[0];    int peak = prices[0];    int maxprofit = 0;    while (i < prices.length - 1) {        while (i < prices.length - 1 && prices[i] >= prices[i + 1])            i++;        valley = prices[i];        while (i < prices.length - 1 && prices[i] <= prices[i + 1])            i++;        peak = prices[i];        maxprofit += peak - valley;    }    return maxprofit;}

复杂度分析：
时间复杂度：O(n)
空间复杂度：O(1)