动态规划实战13 leetcode-121. Best Time to Buy and Sell Stock

前言：这道题目实可以用 Kadane's Algorithm解决

对于kadane's algorithm 我们可以这么想，假设输入是

[7,1,5,3,6,4]

那么我们可以通过这个数组得到在每一天前买后一天卖的数组（称收益数组）

[-6,4,-2,3,-2]

实际上这样就转换成了什么问题？没错！maximum subarray！我们只要求收益数组的maximum subarray!

subarray相当于是持有股票的时间段，比如最大的maximum subarray是 [4,-2,3]，说明我们要在这段时期持有股票

代码

class Solution {public int maxProfit(int[] prices) {//corner caseif(prices==null||prices.length==0||prices.length==1)return 0;int [] nums = new int[prices.length-1];for(int i=1;i<prices.length;i++){nums[i-1]=prices[i]-prices[i-1];}int result= maxSubArray(nums);return result<0 ? 0:result;}public int maxSubArray(int[] nums) {int n = nums.length;int[] dp = new int[n];//dp[i] means the maximum subarray ending with A[i];dp[0] = nums[0];int max = dp[0];for(int i = 1; i < n; i++){dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);max = Math.max(max, dp[i]);}return max;}
}

实际上我们可以将代码简化，可以上leetcode相应solution去寻找。

不过个人感觉这样写的代码更容易让人理解

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