1. Title
Given an array, its i-th element is the price of a given stock on the i-th day.
Design an algorithm to calculate the maximum profit you can get. You can complete up to two transactions.
Note: You cannot participate in multiple transactions at the same time (you must sell the previous stocks before buying again).
Example 1:
输入: [3,3,5,0,0,3,1,4]
输出: 6
解释: 在第 4 天（股票价格 = 0）的时候买入，在第 6 天（股票价格 = 3）的时候卖出，这笔交易所能获得利润 = 3-0 = 3 。
随后，在第 7 天（股票价格 = 1）的时候买入，在第 8 天 （股票价格 = 4）的时候卖出，这笔交易所能获得利润 = 4-1 = 3 。
Example 2:
输入: [1,2,3,4,5]
输出: 4
解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。
因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。
Example 3:
输入: [7,6,4,3,1]
输出: 0
解释: 在这个情况下, 没有交易完成, 所以最大利润为 0。
2. Solve
Series: 【LeetCode】121. The best time to buy and sell stocks , 【LeetCode】122. The best time to buy and sell stocks II
1. Recursion
Ideas:
Examples of two state transitions for buying and selling:
- Do not move/buy --> do not move/sell;
- Do not move/sell-->Do not move/buy
Take the array [1,2,3,4,5] as an example, the recursive call tree is as follows:
Code:
// 参数含义：day：哪一天； status：买入 or 卖出； k：交易次数。
class Solution {
public int maxProfit(int[] prices) {
if (prices==null || prices.length==0) return 0;
int n = prices.length;
return dfs(prices, 0, 0, 0);
}
private int dfs(int[] prices, int day, int status, int k) {
// Terminator：数组执行到头了，或者交易了两次了
if (day==prices.length || k==2) {
return 0;
}
// Current logic：split the problem
//定义三个变量，分别记录[不动]、[买]、[卖]
int hold=0, buy=0, sell=0;
// Drill down: conquer the subproblems
//保持不动
hold = dfs(prices, day+1, status, k);
if (status==1) {
//递归处理卖的情况，这里需要将k+1，表示执行了一次交易
sell = dfs(prices, day+1, 0, k+1) + prices[day];
} else {
//递归处理买的情况
buy = dfs(prices, day+1, 1, k)-prices[day];
}
// Merge the subresults：最终结果就是三个变量中的最大值
return Math.max(Math.max(hold, sell), buy);
}
}
Time complexity: O (2 n) O(2^n)O ( 2n )
Space complexity: O (logn) O(logn)O ( l o g n )
2. Recursion + memoization
Ideas:
The results of repeated calls need to be cached here. In addition, in order to cache the results corresponding to (day-which day; status-buy/sell; k-number of transactions), hashCode() and equals() need to be implemented. This also needs to be understood.
Code:
class Key {
final int day;
final int status;
final int k;
Key(int day, int status, int k) {
this.day = day;
this.status = status;
this.k = k;
}
// 实现自定义hashCode() 与 equals()
public int hashCode() {
return this.day + this.status + this.k;
}
public boolean equals(Object obj) {
Key other = (Key)obj;
if (day==other.day && status==other.status && k==other.k) {
return true;
}
return false;
}
}
class Solution {
public int maxProfit(int[] prices) {
if (prices==null || prices.length==0) return 0;
int n = prices.length;
// 哈希表缓存重复调用
Map<Key, Integer> map = new HashMap<Key, Integer>();
return dfs(map, prices, 0, 0, 0);
}
private int dfs(Map<Key, Integer> map, int[] prices, int day, int status, int k) {
Key key = new Key(day, status, k);
if (map.containsKey(key)) {
return map.get(key);
}
if (day==prices.length || k==2) {
return 0;
}
int hold = 0, sell = 0, buy = 0;
hold = dfs(map, prices, day+1, status, k);
if (status==1) {
sell = dfs(map, prices, day+1, 0, k+1) + prices[day];
} else {
buy = dfs(map, prices, day+1, 1, k) - prices[day];
}
map.put(key, Math.max(Math.max(hold, sell), buy));
return map.get(key);
}
}
Time complexity: O (kn) O(kn)O ( k n ) ?
Space complexity: O (kn) O(kn)O(kn) ?
3. Dynamic planning
Ideas:
1、状态定义：
dp[cnt][i]： 第i天交易cnt次的最大利润。 cnt - 交易次数； i - 天
2、转移方程：
dp[cnt, i] = max(dp[cnt, i-1], prices[i] - prices[j] + dp[cnt-1, j-1]), j=[0..i-1]
3、参考超哥
for day : 0-->n-1
MP[day,0] = max {
MP[day-1, 0], // 不动
MP[day-1, 1]+a[day] // 卖
}
MP[day,1] = max {
MP[day-1,1], // 不动
MP[day-1,0]-a[day] // 买
}
Code:
class Solution {
public int maxProfit(int[] prices) {
if (prices.length == 0) return 0;
int[][] dp = new int[3][prices.length];
for (int k = 1; k <= 2; k++) {
int min = prices[0];
for (int i = 1; i < prices.length; i++) {
min = Math.min(min, prices[i] - dp[k-1][i-1]);
dp[k][i] = Math.max(dp[k][i-1], prices[i] - min);
}
}
return dp[2][prices.length-1];
}
}
Time complexity: O (kn) O(kn)O ( k n ) , k = 3
Space complexity: O (kn) O(kn)O(kn)
Version 2
Ideas:
Same as above, but compress dp, because the value of dp[i] mainly depends on dp[i-1].
Code:
class Solution {
public int maxProfit(int[] prices) {
if (prices.length == 0) return 0;
int[] dp = new int[3];
int[] min = new int[3];
Arrays.fill(min, prices[0]);
for (int i = 1; i < prices.length; i++) {
for (int k = 1; k <= 2; k++) {
min[k] = Math.min(min[k], prices[i] - dp[k-1]);
dp[k] = Math.max(dp[k], prices[i] - min[k]);
}
}
return dp[2];
}
}
Time complexity: O (kn) O(kn)O ( k n ) , k=3
Space complexity: O (k) O(k)O(k)
Version 3
Ideas:
Same as above, readability improved, but versatility decreased.
Code:
class Solution {
public int maxProfit(int[] prices) {
int buy1 = Integer.MAX_VALUE, buy2 = Integer.MAX_VALUE;
int sell1 = 0, sell2 = 0;
for (int i = 0; i < prices.length; i++) {
buy1 = Math.min(buy1, prices[i]);
sell1 = Math.max(sell1, prices[i] - buy1);
buy2 = Math.min(buy2, prices[i] - sell1);
sell2 = Math.max(sell2, prices[i] - buy2);
}
return sell2;
}
}
Time complexity: O (n) O(n)O ( n )
space complexity: O (1) O(1)O ( 1 )
Three, reference
1. Five realizations + detailed illustration 123. The best time to buy and sell stocks III
2. Detail explanation of DP solution