0. 題目
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.
1. c++版本
方法1至壤，使用暴力法凡涩，此方法不能AC，算法復(fù)雜度O(n_2)
int maxProfit(vector<int>& prices) {
int maxProfit = 0;
for (int i=0; i<prices.size(); ++i)
for (int j=i+1; j <prices.size(); ++j) {
if(prices[i] <= prices[j]) {
int currentProfit = prices[j] - prices[i];
maxProfit = max(currentProfit, maxProfit);
}
}
return maxProfit;
}
方法2：只允許一次交易，
思路就是去找到在前一段區(qū)間內(nèi)最小的數(shù)局待，和后一段區(qū)間內(nèi)最大的數(shù)，二者之差即為最大收益报辱。
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.empty())
return 0;
int minValue = prices[0], maxPro = 0;
for (int i=1; i<prices.size(); ++i) {
minValue = min(minValue, prices[i]);
maxPro = max(maxPro, prices[i] - minValue);
}
return maxPro;
}
};
2. python版本
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if not prices:
return 0
minValue, maxPro = prices[0], 0
for i in range(1, len(prices)):
minValue = minValue if minValue < prices[i] else prices[i]
tmpValue = prices[i] - minValue
maxPro = maxPro if maxPro > tmpValue else tmpValue
return maxPro;
