Say you have an array for which the _i_th 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.
思路：
这一题使用dp来做，实质也就是找到当前价之前的最低价，来判断盈利，也可以转换为求最大子数组和（53题）来做，也可以用贪心，直接找出最低价与当前盈利。
代码：
java：
代码语言：javascript复制class Solution {
// 1. time:O(n) space:O(0)
/*public int maxProfit(int[] prices) {
int len = prices.length;
if (len < 1) return 0;
int[] minPrices = new int[len];
int[] maxProfit = new int[len];
minPrices[0] = prices[0];
maxProfit[0] = 0;
for (int i = 1; i < len; i ) {
minPrices[i] = Math.min(minPrices[i-1], prices[i]);
maxProfit[i] = Math.max(maxProfit[i-1], prices[i] - minPrices[i-1]);
}
return maxProfit[len-1];
}*/
// 2. time: O(n) space:O(1)
public int maxProfit(int[] prices) {
int len = prices.length;
if (len < 1) return 0;
int profit = 0;
int buyPrice = prices[0];
for (int i = 1; i < len; i ) {
buyPrice = Math.min(prices[i], buyPrice);
profit = Math.max(prices[i] - buyPrice, profit);
}
return profit;
}
}
