Leetcode Best Time to Buy and Sell Stock III

Leetcode Best Time to Buy and Sell Stock III，本算法的关键为找出其动态子结构。可以发现，序列中的最小值可以做为其的一个分割，令左边序列为left，右边的序列为right，整个串为whole，可以推出当前串的最大值一定是在：a.左序列一次transaction最大值(left_max)与右序列一次transaction最大值(right_max)之和，b.整个序列一次交易最大值；两者之间产生。有一种情况.

class Solution {public:    int maxProfit(vector<int>& prices) {        int max = 0;        findMax(prices, max, 0, prices.size() - 1);        return max;    }    int findMax(vector<int>& prices, int& max, int start, int end) {        if (start >= end) {            return 0;        }        int begin = start;        int cur_max = 0;        int left_max = 0;        int right_max = 0;        int min_pos = -1;        int pre = start;        int descent_count = 0;        int backward = 0;        for (int i = start + 1; i <= end; i ++) {            // get the consective descent element num            if (prices[i] <= prices[pre]) {                descent_count ++;            } else {                descent_count = 0;            }            // get the minest element            if ((prices[i] <= prices[min_pos] || min_pos == -1) &&                    prices[i] < prices[pre]) {                min_pos = i;                backward = descent_count;            }            // calculate the max profite by one transaction            if (prices[i] < prices[begin]) {                begin = i;            } else {                int tmp = prices[i] - prices[begin];                cur_max = tmp > cur_max? tmp : cur_max;            }            pre = i;        }        // calculate the left and the right        if (min_pos != -1) {            left_max = findMax(prices, max, start, min_pos - backward);            right_max = findMax(prices, max, min_pos, end);        }        // get the current max        if (cur_max > max) {            max = cur_max;        }        if (left_max + right_max > max) {            max = left_max + right_max;        }        return cur_max;    }};

test: ./a.out 3 2 4 2 5 7 2 4 9 0re: 12