题目描述
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.
Example1:
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.
Example2:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
解题思路
- 只需要遍历一次数组，用一个变量记录遍历过数中的最小值，
- 然后每次计算当前值和这个最小值之间的差值最为利润，然后每次选较大的利润来更新。
- 当遍历完成后当前利润即为所求
class Solution {
public:
int maxProfit(vector<int>& prices) {
int buy = INT_MAX;
int res = 0;
for(int i = 0; i < prices.size(); i++){
buy = min(buy, prices[i]);
res = max(res, prices[i] - buy);
}
return res;
}
};
第二遍出现的问题
- 思路错误，总是会想到把结果累加
- 应该直接将最大的与最小的进行相比，然后更新结果
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() == 0){
return 0;
}
int buy = prices[0];
int res = 0;
for(int i = 1; i < prices.size(); i++){
buy = min(buy, prices[i]);
res = max(res, prices[i] - buy);
}
return res;
}
};
