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.
1.思考
- 这个虽然是I的加强版，但是修改了条件之后解题思路就完全不一样了；
- 通过将prices用折线图画出来，可以想到本题的解题思路是这样的：
(1) 先将所有上升的价格进行标注；
(2) 再将所有的上升价格起始、中止以及差值都记录下来；
(3) 然后对每一段相邻的上升区间进行比较，若有重叠，则分段买入和抛售，对差值直接相加；若没有重叠，则在前一段初值买入，后一段中止值抛售，两者相减。 - 这个方法的空间复杂度是较高的，还有待优化。
2.实现
Runtime: 8ms (99.42%)
Memory: 9.8MB (5.39%)
class Solution {
public:
int maxProfit(vector<int>& prices) {
int len = prices.size();
vector<int> fup(len, -1);
for(int i=0; i<len-1; i++){
if(prices[i]<prices[i+1])
fup[i] = 1;//The price will increase.
}
vector<vector<int>> incr;
int j=0, f, b;
while(j<len-1){
if(fup[j]==1){
f = prices[j];
while(fup[j]==1)
j++;
b = prices[j];
incr.push_back(vector<int> {f, b, b-f});
}
j++;
}
int li = incr.size();
if(li==0)
return 0;
if(li==1)
return incr[0][2];
int pro = incr[0][2];
for(int i=0; i<li-1; i++){
auto pi = incr[i];
auto pj = incr[i+1];
if(pi[1]>=pj[0])
pro += pj[2];
else
pro += pj[1] - pi[0] - pi[2];
}
return pro;
}
};