Best Time to Buy and Sell Stock IV
Desicription
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 at most k transactions.
Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Example 1:
代码语言：javascript复制Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.Example 2:
代码语言：javascript复制Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.Solution
代码语言：javascript复制class Solution {
public:
int maxProfit(int k, vector<int>& prices) {
priority_queue<int> profits;
stack<pair<int, int>> vps;
int v = 0;
int p = 0;
int len = prices.size();
while(p < len) {
for(v = p; v < len-1 && prices[v] >= prices[v 1]; v );
for(p = v 1; p < len && prices[p] >= prices[p-1]; p );
while(vps.size() && prices[vps.top().first] > prices[v]) {
profits.push(prices[vps.top().second - 1] - prices[vps.top().first]);
vps.pop();
}
while(vps.size() && prices[vps.top().second-1] <= prices[p-1]) {
profits.push(prices[vps.top().second-1] - prices[v]);
v = vps.top().first;
vps.pop();
}
vps.push(pair<int, int>(v, p));
}
while(vps.size()) {
profits.push(prices[vps.top().second-1] - prices[vps.top().first]);
vps.pop();
}
int res = 0;
while(k-- && profits.size())
res = profits.top(), profits.pop();
return res;
}
};
