LeetCode_121. Best Time to Buy and Sell Stock

121. Best Time to Buy and Sell Stock

Easy

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

package leetcode.easy;

public class BestTimeToBuyAndSellStock {
	@org.junit.Test
	public void test() {
		int[] prices1 = { 7, 1, 5, 3, 6, 4 };
		int[] prices2 = { 7, 6, 4, 3, 1 };
		System.out.println(maxProfit1(prices1));
		System.out.println(maxProfit1(prices2));
		System.out.println(maxProfit2(prices1));
		System.out.println(maxProfit2(prices2));
	}

	public int maxProfit1(int[] prices) {
		int maxprofit = 0;
		for (int i = 0; i < prices.length - 1; i++) {
			for (int j = i + 1; j < prices.length; j++) {
				int profit = prices[j] - prices[i];
				if (profit > maxprofit) {
					maxprofit = profit;
				}
			}
		}
		return maxprofit;
	}

	public int maxProfit2(int[] prices) {
		int minprice = Integer.MAX_VALUE;
		int maxprofit = 0;
		for (int i = 0; i < prices.length; i++) {
			if (prices[i] < minprice) {
				minprice = prices[i];
			} else if (prices[i] - minprice > maxprofit) {
				maxprofit = prices[i] - minprice;
			}
		}
		return maxprofit;
	}
}