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 numif (prices[i] <= prices[pre]) {descent_count ++;} else {descent_count = 0;}// get the minest elementif ((prices[i] <= prices[min_pos] || min_pos == -1) &&prices[i] < prices[pre]) {min_pos = i;backward = descent_count;}// calculate the max profite by one transactionif (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 rightif (min_pos != -1) {left_max = findMax(prices, max, start, min_pos - backward);right_max = findMax(prices, max, min_pos, end);}// get the current maxif (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 0
re: 12

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