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剑指 Offer 63. 股票的最大利润

思路

方法一：暴力法

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         int n = (int)prices.size(), ans = 0;
 5         for (int i = 0; i < n; ++i){
 6             for (int j = i + 1; j < n; ++j) {
 7                 ans = max(ans, prices[j] - prices[i]);
 8             }
 9         }
10         return ans;
11     }
12 };

 

方法二：动态规划

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         if(prices.size() < 2)    //这种情况无法交易
 5             return 0;
 6             
 7         vector<int> dp(prices.size(), 0);
 8         int minPrice = prices[0];
 9         dp[0] = 0;
10 
11         for(int i = 1; i < prices.size(); ++i) {
12             if(prices[i] < minPrice)
13                 minPrice = prices[i];
14             
15             dp[i] = max(dp[i-1], prices[i] - minPrice);
16         }
17 
18         return dp[prices.size()-1];
19     }
20 }

复杂度分析

时间复杂度：O(n)

空间复杂度：O(n)

空间优化

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         if(prices.size() < 2)
 5             return 0;
 6 
 7         int minPrice = prices[0];
 8         int profit = 0;
 9 
10         for(int i = 1; i < prices.size(); ++i) {
11             if(prices[i] < minPrice)
12                 minPrice = prices[i];
13             
14             profit = max(profit, prices[i] - minPrice);
15         }
16 
17         return profit;
18     }
19 };

复杂度分析

时间复杂度：O(n)

空间复杂度：O(1)

 

posted @ 2020-11-15 19:12  拾月凄辰  阅读(101)  评论(0)  编辑  收藏  举报