【LeetCode】Best Time to Buy and Sell Stock III

https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

这题没做出来，看了网上的分析才做出来的。。。关键点就是要利用on-line算法的性质，避免重复计算。

class Solution {public:    int maxProfit(vector<int> &prices) {        if (prices.size() < 2) {            return 0;        }        vector<int> max_profile1; max_profile1.reserve(prices.size() + 1); // max_profile1[i] : max profit of [0, i]        vector<int> max_profile2; max_profile2.reserve(prices.size() + 1); // max_profile2[i] : max profit of [i, end];                max_profile1[0] = 0;        int low = prices[0], temp = 0;        for (int i = 1, end = prices.size(); i < end; ++i) {            if (prices[i] < low ) { low = prices[i]; }            if (prices[i] - low > temp) { temp = prices[i] - low; }            max_profile1[i] = temp;        }                max_profile2[prices.size()] = 0;        int hi = prices[prices.size()-1]; temp = 0;         for (int i = prices.size() - 1; i >= 0; --i) {            if (hi < prices[i]) { hi = prices[i]; }            if (temp < hi - prices[i]) { temp = hi - prices[i]; }            max_profile2[i] = temp;        }                int max_profile = 0;        for (vector<int>::size_type i = 0, end = prices.size(); i < end; ++i) {            if (max_profile < max_profile1[i] + max_profile2[i+1]) {                max_profile = max_profile1[i] + max_profile2[i+1];            }        }        return max_profile;    }};