Problem
Best Time to Buy and Sell Stock II
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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
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: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 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
This problem is to follow all the risings and trade the upslope. The simplest solution is to add up all the upturns.
public class BuyAndSellStockII {
public static int gain(int[] a) {
int N = a.length;//5
int gain = 0;
if(N <= 1) return 0;
for(int i = 1; i < N; i++) {//0
int delta = a[i] - a[i-1];
if(delta > 0) {
gain += delta;
}
}
return gain;
}
public static void main(String...args) {
System.out.println(gain(new int[]{7,1,5,3,6,4}));
System.out.println(gain(new int[]{1,2,3,4,5}));
System.out.println(gain(new int[]{7,6,4,3,1}));
}
}
public static int gain(int[] a) {
int N = a.length;//5
int gain = 0;
if(N <= 1) return 0;
for(int i = 1; i < N; i++) {//0
int delta = a[i] - a[i-1];
if(delta > 0) {
gain += delta;
}
}
return gain;
}
public static void main(String...args) {
System.out.println(gain(new int[]{7,1,5,3,6,4}));
System.out.println(gain(new int[]{1,2,3,4,5}));
System.out.println(gain(new int[]{7,6,4,3,1}));
}
}
Time cost O(N), space cost O(1).
