Leetcode_best-time-to-buy-and-sell-stock-ii(c++ and python version)

地址： http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思路：找拐点，波峰的概念，每次交易都是 波峰high - 波谷low。解释的有点抽象，参考代码想一想理解一下。

参考代码：

class Solution {
public:
    int maxProfit(vector &prices) {
        if(prices.empty())
            return 0;
        int res = 0, low, high;
        for(int i = 0; i < prices.size()-1; ++i)
        {
            if(prices[i]

网上看到一份代码，如下

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

解法很清楚明了，但是不符合交易的概念，因为存在卖出去同时又马上买进来的股票，如[ 1 , 3, 5]

python参考代码：

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        if not prices or len(prices)==1: return 0
        ans = 0
        begin = prices[0]
        for i in range(1, len(prices)-1):
            if begin > prices[i]:
                begin = prices[i]
            if prices[i] > prices[i+1]:
                ans += prices[i] - begin
                begin = prices[i+1]
        if prices[-1] > begin:
            ans+=prices[-1] - begin;
        return ans