题目描述

• 讲道理，一眼dp
  

代码 & 思路

• 时间复杂度O(n)，不过可改进的地方还多，跑出来大概6ms。

初版代码

class Solution {public int maxProfit(int[] prices) {// 做啥：找一个i,j（j > i)且prices[j] > prices[i]，使得 max = prices[j] - prices[i]// 用dp做吧int len = prices.length;if(len == 1){return 0;}// dp[i]代表当前值之后能遇到的最大值int[] dp = new int[len];dp[len-1] = 0;for(int i=len-2;i >= 0;i--){dp[i] = Math.max(prices[i+1],dp[i+1]);}int max = 0;for(int i=0;i<len-1;i++){max = Math.max(dp[i] - prices[i],max);}return max;}
}

更新啦～优化代码

• 节约了空间，空间复杂度由 O(n) 变成了 O(1)

class Solution {public int maxProfit(int[] prices) {if(prices.length < 2) {return 0;}int len = prices.length;int max = prices[len - 1];int ans = 0;for(int i = len - 2; i >= 0; i--) {int nowProfit = max - prices[i];// ans 更新if(nowProfit > ans) {ans = nowProfit;}// min 更新if(prices[i] > max) {max = prices[i];} }return ans;}
}

再次更新

class Solution {public int maxProfit(int[] prices) {int max = prices[prices.length - 1];int res = 0;for(int i = prices.length - 2; i >= 0; i--) {max = Math.max(max, prices[i + 1]);res = Math.max(max - prices[i], res);}return res;}
}