[LeetCode]121. Best Time to Buy and Sell Stock 解题报告(C++)

[LeetCode]121. Best Time to Buy and Sell Stock 解题报告(C++)

题目描述

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.

题目大意

• day i 的价格为 arr[i].
• 仅仅通过一次交易.让收益最大.

解题思路

方法1:

• 暴力方法:

• 双重循环遍历

代码实现:

class Solution0{
public:
    int maxProfit(vector<int>& prices) {
        int size = prices.size();
        int res = 0;
        // 最多一次交易
        for (int i = 0; i < size; i++) {
            for (int j = i + 1; j < size; j++) {
                if (prices[j] > prices[i]) {
                    res = max(res, prices[j] - prices[i]);
                }
            }
        }
        return res;
    }
};

方法2:

• 用变量记录 前面遍历过的数字的最小值,
• 再比较当前价格与最小值的差值和当前最佳利润大小..
• 更新利润

代码实现:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        // 最多一次交易
        int buy = INT_MAX;
        for(int price:prices){
            buy = min(price,buy);
            res = max(price-buy,res);
        }
        return res;
    }
};

小结

• 由于只有一次交易,所以,是找到前面的最小和后面最大做差.
• 方法2的用 min 和 max 做更新 实在是太厉害了.