122. Best Time to Buy and Sell Stock II (Array)

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.
##### 分析：

和另外一道类似，但是不同的在于可以多次买卖，但是手中同时进行的交易不能大于1个。

###### 方法一：

用max保存最大值，min保存最小值，res保存临时收益，profit保存最终收益，当当前价格比max小时应该结束当前交易，从当前价格买入，然后更新min，max都是为当前价格，temp加入profit,temp归零；如果当前价格比max大时，继续当前交易，更新max。时间复杂度O(n),空间复杂度O(1)

public int maxProfit(int[] prices) {
if(prices==null ||prices.length==0)
return 0;
int min=prices[0], max=prices[0];
int profit=0;
int temp = max-min;
for(int i=1;i<prices.length;i++){
if(prices[i]<max){
temp = max-min;
profit += temp;
temp = 0;
min = prices[i];
max = prices[i];
}
else{
max = prices[i];
temp = max-min;
max = prices[i];
}
}
profit += temp;
return profit;
}
###### 方法二：Peak Valley Approach

TotalProfit=∑i​(height(peaki​)−height(valleyi​))

和方法一差不多，在一个while循环中首先找valley，遇到升高就开始找peak.再次降低结束这次循环。继续开始搜索下一个上坡。

Time complexity : O(n)O(n). Single pass.

Space complexity : O(1)O(1). Constant space required.

public int maxProfit(int[] prices) {
int i = 0;
int valley www.dasheng178.com= prices[0];
int peak = prices[0];
int maxprofit = 0;
while (i < prices.length - 1) {
while (i < prices.length - 1 && prices[www.xiaomiyulezc.com ] >= 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;
}
方法三：

只关注上坡，然后将每一段的profit相加

Time complexity : O(n)O(n). Single pass.

Space complexity : O(1)O(1). Constant space required.

public int maxProfit(int[] prices) {
int maxprofit www.michenggw.com= 0;
for (int i = 1; i <www.mhylpt.com/ prices.length; i++) {
if (prices[i] > prices[i - 1])
maxprofit += prices[i] - prices[i - 1];
}
return maxprofit;