题目意思:
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.
大意就是:
给定一个整型数组,代表股票的波动价格。假设你最多只能做一次交易,如何使自己的收益最大化?
做题思路:
1,最容易想到的是嵌套循环,但时间复杂度要n^2,超时了。
示例代码如下:
class Solution{
public:
int maxProfit(vector<int> &prices){
vector<int>::size_type length = prices.size();
if(length == 0){
return 0;
}
int i, j, max = 0;
for(i = 0;i < length-1;i++){
int sell_price = 0;
if((i >0) && (prices[i] < prices[i-1])){
for(j=i+1;j<length;j++){
if(prices[j] > prices[j-1]){
int temp = prices[j] - prices[i];
if(max < temp){
max = temp;
}
}
}
}
}
return max;
}
};
2,因此可以采用一种改进的思路。只需要一重循环:当前位置之前选出最小的买入价格,当前位置之后计算最大的利润。
代码如下:
class Solution{
public:
int maxProfit(vector<int> &prices){
vector<int>::size_type length = prices.size();
if(length == 0){
return 0;
}
int i, profit = 0, min = prices[0];
for(i = 1;i<length;i++){
profit = (prices[i] - min > profit ? prices[i] - min : profit);
min = (prices[i] < min ? prices[i] : min);
}
return profit;
}
};