122. Best Time to Buy and Sell Stock II

标签：tco   多次   pre   mission   因此   must   tip   leetcode   ++   

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

给定一个数组，其中数组的第 i 个元素表示第 i 天的股票价格，你可以选择在其中一天买股票，然后再后面的某一天卖掉股票，求最大收益。（股票必须要先买才能卖）这里可以购买多次，也可以卖出多次
-------------------
Example 1:
Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
-----------------
Example 2:
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
---------------------
Example 3:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

分析：

与 https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ 类似，解答见https://www.jianshu.com/p/cdb6881094f0，区别在于这里可以买卖多次

法一

我们可以依次找到局部最小值，然后再在局部最小值后面找到局部最大值，在这两个地方买进和卖出；然后再依次往后找：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        
        int buy = 0, sell = 0;
        int i = 0, len = prices.size()-1;
        int profit = 0;
        while(i < len)
        {
            while(i < len && prices[i+1] <= prices[i]) // find the local minimum
                ++i;
            buy = prices[i]; // buy at the local minimum
            
            while(i < len && prices[i+1] >= prices[i]) // find the local maximum
                ++i;
            sell = prices[i]; // sell at the local maximum
            
            profit += (sell-buy);
        }
        return profit;
    }
};

结果：

Runtime: 4 ms
Memory Usage: 7.4 MB

方法2

从 121. Best Time to Buy and Sell Stock，我们可以知道，
nums[j] - nums[i] = nums[j] - num[j-1] + nums[j-2] - nums[j-3] + .... + nums[i+1] - nums[i]
方法一也是找到一个局部递增子数组，因此可以直接判断 如果 nums[i+1] > nums[i]，就可以直接买进和卖出

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int profit = 0;
        for(int i = 1; i<prices.size(); ++i)
        {
            if(prices[i] > prices[i-1])
            {
                profit += (prices[i]-prices[i-1]);
            }
        }
        return profit;
    }
};

结果

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Best Time to Buy and Sell Stock II.
Memory Usage: 7.4 MB, less than 100.00% of C++ online submissions for Best Time to Buy and Sell Stock II.

122. Best Time to Buy and Sell Stock II

标签：tco   多次   pre   mission   因此   must   tip   leetcode   ++   

原文地址：https://www.cnblogs.com/qiulinzhang/p/12639999.html