LeetCode - Best Time to Buy and Sell Stock II （贪心对策，差分序列）
LeetCode -- Best Time to Buy and Sell Stock II （贪心策略，差分序列）
Best Time to Buy and Sell Stock II
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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).
解题思路：
贪心策略
分析一：
对于一个股票的价格走势，多次买入卖出使收益最大。目标就是找出这个股市波动曲线中一段段的上升曲线，对于每一个上升曲线，计算其收益，最后受益相加。（详见下面Java源码）
分析二：
同样对于这个波动曲线，低进高出，每一天间隔划分为相应的收益段，把所有的正的价格差相加起来。
实现通过： 把原始价格序列变成差分序列（见下面C++源码）
源代码：
分析一：Java源代码
public static int maxProfit(int[] prices) {
if (prices.length == 0)
return 0;
int i = 0;
int profit = 0;
int begMin = prices[0];
for (i = 1; i < prices.length; ++i) {
if (prices[i] < prices[i - 1]) {
profit += prices[i - 1] - begMin;
begMin = prices[i];
}else if (i == prices.length-1){
profit += prices[i] - begMin;
}
begMin = Math.min(begMin, prices[i]);
}
return profit;
}
分析二： C++源代码
int maxProfit(vector<int> &prices) {
int sum=0;
for(int i=1; i<prices.size(); i++){
int diff = prices[i]-prices[i-1];
if( diff>0 )
sum += diff;
}
return sum;
}