这道题我一开始想到用递归方法，可以把规模大的问题变成规模小的问题，但是觉得递归的时间复杂度很高，因为它会把相同的问题进行重复计算，然后我想是不是有什么down-up的方法，先把所有的子问题的结果保存起来，但是发现问题的最优解并不能由子问题的最优解推导出来。最后就想到买股票的时候，我们在一个局部极小的时候买进，然后在一个局部极大的时候卖出，这样就可以通过多次买进卖出交易获得最大收益。接下来的问题就是，确定什么时候是极小值，什么时候是极大值。
下面是AC代码：
1 /**
2 * Design an algorithm to find the maximum profit. You may complete as many transactions as you like
3 * (ie, buy one and sell one share of the stock multiple times).
4 * However, you may not engage in multiple transactions at the same time
5 * (ie, you must sell the stock before you buy again).
6 * @param prices
7 * @return
8 */
9 public int maxProfit2(int[] prices){
10 if(prices==null || prices.length <= 1)
11 return 0;
12 int profit = 0;
13 int len = prices.length;
14 //the local minimum is marked as -1;the local maximum is marked as 1; otherwise is 0
15 int[] mark = new int[len];
16 int[] left = new int[len];//mark whether the left is <= or >=
17 int[] right = new int[len];//mark whether the right is <= or >=
18 // the first number is initialized as 0
19 left[0] = 0;
20 //get the left array, by one traversal
21 int i=1;
22 while(i<len){
23 if(prices[i-1]<prices[i])
24 left[i] = -1;
25 else if(prices[i-1]>prices[i])
26 left[i] = 1;
27 else
28 left[i] = left[i-1];
29 i++;
30 }
31 //initialize the last element
32 right[len-1] = 0;
33 i = len-2;
34 while(i>=0){
35 if(prices[i+1]<prices[i])
36 right[i] = -1;
37 else if(prices[i+1]>prices[i])
38 right[i] = 1;
39 else
40 right[i] = right[i+1];
41 i--;
42 }
43 //get the mark
44 i = 0;
45 while(i<len){
46 if(left[i] == right[i]){
47 mark[i] = left[i]*(-1);
48 }else {
49 if(left[i] == 0){
50 mark[i] = right[i]*(-1);
51 }
52 else if(right[i] == 0)
53 mark[i] = left[i]*(-1);
54 else
55 mark[i] = 0;
56 }
57 i++;
58 }
59 int buy=0 ,sell = 0;
60 boolean f1 = false, f2 = false;
61 for(i=0;i<len;i++){
62 if(f1==false && mark[i] == -1)//at the minimum time buy.
63 {
64 buy = prices[i];
65 f1 = true;
66 }
67 if(f2 == false && f1 == true && mark[i] == 1)//at the maximum time sell, by the condition that we have bought the stock;
68 {
69 sell = prices[i];
70 f2 = true;
71 }
72 if(f1 && f2)
73 {
74 profit += (sell - buy);
75 f1 = false;
76 f2 = false;
77 }
78 }
79 return profit;
80 }
LeetCode - Best Time to Buy and Sell 2,码迷,mamicode.com
时间： 2024-08-08 09:29:43