标签：解题思路 leetcode public sel tco pac += int span
一：解题思路
方法一：不断的寻找股票的最小值和最大值，并在最小值得时候买入，在最大值得时候卖出。Time:O(n),Space:O(1)
方法二：贪心法，只要后一天的值大于前一天的值，那么就进行买卖。Time:O(n),Space:O(n)
二：完整代码示例 (C++版和Java版)
方法一C++：
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.size() == 0) return 0; int n = prices.size(); int i = 0, buy = 0, sell = 0,maxProfit=0; while (i < n - 1) { while (i < n - 1 && prices[i + 1] <= prices[i]) i++; buy = prices[i]; while (i < n - 1 && prices[i + 1] >= prices[i]) i++; sell = prices[i]; maxProfit += (sell-buy); } return maxProfit; } };
方法一Java：
class Solution { public int maxProfit(int[] prices) { if(prices==null||prices.length==0) return 0; int maxProfit=0,i=0,buy=0,sell=0; int n=prices.length; while(i<n-1) { while(i<n-1 && prices[i+1]<=prices[i]) i++; buy=prices[i]; while(i<n-1 && prices[i+1]>=prices[i]) i++; sell=prices[i]; maxProfit+=(sell-buy); } return maxProfit; } }
方法二C++:
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.size() == 0) return 0; int maxProfit=0; for (int i = 1; i < prices.size(); i++) { if (prices[i] > prices[i - 1]) maxProfit += (prices[i]-prices[i-1]); } return maxProfit; } };
方法二Java:
class Solution { public int maxProfit(int[] prices) { if(prices==null||prices.length==0) return 0; int maxProfit=0; for(int i=1;i<prices.length;i++) { if(prices[i]>prices[i-1]) maxProfit+=(prices[i]-prices[i-1]); } return maxProfit; } }
p64 不限次数进行股票买卖的最大利润 (leetcode 122)
标签：解题思路 leetcode public sel tco pac += int span
原文地址：https://www.cnblogs.com/repinkply/p/12527154.html
