- Best Time to Buy and Sell Stock III
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 1:
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Solution
Approach1 以 i 为分界,左边是第一次交易能够获得的最大利润,右边是第二次,最后两边加起来取最大
Note:不能每次都计算一次,用数组存储能够获得的最大利润,否则会超时
class Solution {
public:
int maxProfit(vector<int>& prices) {
int ans = 0;
int n = prices.size();
if(n == 0)return ans;
int left[n] = {0}, right[n] = {0};
int min_price = prices[0];
for(int i = 1; i < n; ++i){
min_price = min(min_price, prices[i]);
left[i] = max(left[i-1], prices[i] - min_price);
}
int max_price = prices[n-1];
for(int i = n-2; i >= 0; --i){
max_price = max(max_price, prices[i]);
right[i] = max(right[i+1], max_price - prices[i]);
}
for(int i = 0; i < n; ++i){
ans = max(ans, left[i] + right[i]);
}
return ans;
}
};
Appraoch 2 每次取最值时针对全局的利润,设置4个变量:b1, s1, b2, s2
class Solution {
public:
int maxProfit(vector<int>& prices) {
int b1 = INT_MIN, b2 = INT_MIN;
int s1 = 0, s2 = 0;
for(int x : prices){
b1=max(b1,-x); //以低价买入
s1=max(s1,b1+x); //以高价卖出
b2=max(b2,s1-x); //低价买入,即结余要最大
s2=max(s2,b2+x); //高价卖出
}
return s2;
}
};