122. Best Time to Buy and Sell Stock II

题目：

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 (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

算法

贪心思想，最大的收益方法就是尽可能多的低入高抛，只要明天比今天价格高，就今天买，明天卖
如图：每个邻近的极大值(peak)和极小值(valley)之间的差相加，一定是最大的。
当然，也可以不判断极大极小，从头到尾遍历一遍，只要后一个比前一个大，就累加上去，思想是一样的。
时间复杂度都是O(n).

代码

int maxProfit(int* prices, int pricesSize) {
    if(pricesSize==0||pricesSize==1){
        return 0;
    }
    int i;
    int min = -1;
    if(prices[0]<=prices[1]){
        min = prices[0];
    }//valley
    
    int profit = 0;
    for( i =1;i<pricesSize-1;i++){
        if(prices[i]<=prices[i-1]&&prices[i]<=prices[i+1]&&min==-1){
            min = prices[i];
        }//valley
        
        if(prices[i]>=prices[i-1]&&prices[i]>=prices[i+1]&&min!=-1){
            profit+=(prices[i]-min);
            min=-1;
        }//peak       
    }
    
    if(min!=-1&&prices[pricesSize-1]>=prices[pricesSize-2]){
        profit+=(prices[i]-min);
    }//peak 
    
    return profit;
}