121. Best Time to Buy and Sell Stock | 买股票的利润

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

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思路: 遍历数组，记录当前已经遍历的价格的最小值，如果当前遍历的价格大于其前面的最小值，则减去最小值，如果利润大于最大利润，则记录最新的最大利润。其实就是比较了，所有数与其之前的最小价格相减的值，最大的就是利润最大值。

          之前一直在找用两个整数进行同时进行遍历的方法，但是会超时，要找到的规律是如果在当前天卖掉的话，想获得最利润就是当天的价格减去之前的最小值，其实这种方法也可倒着来，就是在当天买的话，要获得最大利润就需要用它之后的最大值减去它。

public class Solution {
   public int maxProfit(int[] prices) {
		if (prices.length == 0) {
			return 0;
		}
		int profit = 0;

		int min = prices[0];
		for (int i = 1; i < prices.length; i++) {
			if (prices[i] < min) {
				min = prices[i];
			} else {
				if (prices[i] - min > profit) {
					profit = prices[i] - min;
				}
			}
		}

		return profit;
	}
}