题目描述

已知一个数组,第i个元素表示第i天股票的价格,你只能进行一次交易(买卖各一次),设计算法找出最大收益

测试样例

Input: [7, 1, 5, 3, 6, 4]
Output: 5
最大收益 = 6-1 = 5 (不是7-1 = 6,因为先买后卖,7买,1买亏了6) Input: [7, 6, 4, 3, 1]
Output: 0
最大收益为0

详细分析

初看非常简单,遍历数组,每次选择一个元素,找到这个元素后面的数组的最大值,计算差值,和当前最大收益比较即可,就像这样:

[7,1,5,3,6,4] 当前7,后面最大6,收益-1

[7,1,5,3,6,4] 当前1,后面最大6,收益5

[7,1,5,3,6,4] 当前5,后面最大6,收益1

如此继续找到即可。

不过这种方法会超时,稍微改变一下,现在不止保持最大收益,还保存最低价格,如果当天价格更低,就刷新最小价格(买),同时如果当天价格减去最小价格的收益最大,就刷新最大价格(卖),过程如下:

[7,1,5,3,6,4] minPrice=INT_MAX,maxProfit=0 => minPrice=7,maxProfit=0

[7,1,5,3,6,4] minPrice=7,maxProfit=0 => minPrice=1,maxProfit=0

[7,1,5,3,6,4] minPrice=1,maxProfit=0 => minPrice=1,maxProfit=4

[7,1,5,3,6,4] minPrice=1,maxProfit=4 => minPrice=1,maxProfit=4

[7,1,5,3,6,4] minPrice=1,maxProfit=4 => minPrice=1,maxProfit=5

[7,1,5,3,6,4] minPrice=1,maxProfit=5 => minPrice=1,maxProfit=5

算法实现

  • 方法1:Time Limit Exceeded
class Solution {
public:
int maxProfit(vector<int>& prices) {
int profit = 0;
for(auto start=prices.begin();start!=prices.end();start++){
int buy = *start;
int sell = *std::max_element(start,prices.end());
if((sell - buy)>profit){
profit = sell-buy;
}
}
return profit;
}
};
  • 方法2: Accepted
class Solution{
public:
int maxProfit(vector<int> &prices){
int profit = 0;
int minBuy = std::numeric_limits<int>::max();
for(int i=0;i<prices.size();i++){
//buy it if current price is less than minimum prices yet
if(prices[i]<minBuy){
minBuy = prices[i];
}
//sell it if current profit got optimally
if((prices[i]-minBuy)>profit){
profit = prices[i]-minBuy;
}
}
return profit;
}
};

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