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如果不限交易次数,把所有递增序列差值求和即可。
代码:
1 int maxProfit(vector<int> &prices) { 2 if (prices.empty()) 3 return 0; 4 5 int profit = 0; 6 int climax = prices[prices.size() - 1]; 7 8 for (int i = prices.size() - 2; i >= 0; i--) { 9 if (prices[i] >= prices[i + 1]) { 10 profit += climax - prices[i + 1]; 11 climax = prices[i]; 12 } 13 } 14 profit += climax - prices[0]; 15 16 return profit; 17 }
Leetcode#122 Best Time to Buy and Sell Stock II
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原文地址:http://www.cnblogs.com/boring09/p/4262487.html