Leetcode Best Time to Buy and Sell Stock IV(最大子段和)

朴素的O(n^2)过不了，这里借鉴前人的思想，总结一下。

主要利用栈维护对后面产生影响的字段，然后利用贪心的思想逐步合并。

首先，找到每一对最低点(vx)，最高点(px);

其次，合并当前(vx, px) 与栈内的点对(top_vx, top_px)，分两种情况

1、若top_vx >= vx，考虑到vx处在更低点，(top_vx, top_px)对后面的点对不会有影响，则对(top_vx, top_px)计算价值并出栈；

2、若top_px <= px(ps:此时top_vx < vx),

a、若最优解只包含两对点的其中一对，显然可以合并两对点为(top_vx, px)最优；

b、若最优解中包含两对点，则可将两对点转化成(top_vx, px), (vx, top_ px), 其中(vx, top_ px)对后面的点对不会有影响，则对(vx, top_ px)对计算价值并出栈，同时更新(vx, px)->(top_vx, px)；

最后，将栈内的点对全部计算价值(显然栈内的点对都不必合并)。

class Solution {stack< pair<int, int> > waiting;vector<int> profits;
public:int maxProfit(int k, vector<int>& pr) {int len = pr.size();if(len < 2 || k < 1) return 0;while(!waiting.empty()) waiting.pop();profits.clear();int posv, posp = -1;while(true){for(posv = posp + 1; (posv + 1) < len && pr[posv] >= pr[posv + 1]; posv++);for(posp = posv + 1; (posp + 1) < len && pr[posp] <= pr[posp + 1]; posp++);if(posp >= len) break;for(; !waiting.empty() && pr[waiting.top().first] >= pr[posv]; waiting.pop()){profits.push_back(pr[waiting.top().second] - pr[waiting.top().first]);}for(; !waiting.empty() && pr[waiting.top().second] <= pr[posp]; waiting.pop()){profits.push_back(pr[waiting.top().second] - pr[posv]);posv = waiting.top().first;}waiting.push(make_pair(posv, posp));}for(; !waiting.empty(); waiting.pop()){profits.push_back(pr[waiting.top().second] - pr[waiting.top().first]);}int ans = 0;if(k >= profits.size()) {ans = accumulate(profits.begin(), profits.end(), 0);}else{nth_element(profits.begin(), profits.begin() + k - 1, profits.end(), greater<int>());ans = accumulate(profits.begin(), profits.begin() + k, 0);}return ans;}
};

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