27. Best Time to Buy and Sell Stock && Best Time to Buy and Sell Stock II && Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock

（onlineJudge: https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock/）

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

注意： 限制条件： 先买后卖（不同天）。

思想： 买了后，1. 若以后价格不变，不买不卖。 1. 更价格低，重新买。2. 价格升高，假定抛售，更新一下利润值。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int buy = 0x7fffffff, maxProfile = 0;
        for(int i = 0; i < prices.size(); ++i) {
            if(prices[i] == buy) continue;
            if(prices[i] < buy) { buy = prices[i]; continue; }
            else maxProfile = max(maxProfile, prices[i]-buy);
        }
        return maxProfile;
    }
};

Best Time to Buy and Sell Stock II

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

Say you have an array for which the i^th 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).

思想：求出所有非递减序列两端绝对值之和。我贴在 leedcode 的代码和证明:

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n = prices.size();
        if(n == 0) return 0;
        int start = 0, profile = 0;
        for(int i = 1; i < n; ++i) {
            if(prices[i] < prices[i-1]) {
                profile += prices[i-1] - prices[start];
                start = i;
            }
        }
        profile += prices[n-1] - prices[start];
        return profile;
    }
};
/*********************** *provement ***************/
/*Explain the code I pasted above: 
 
From left to right I find out every subsequence that not exist decrease. 
 
such as: l ... k1 ...k2 ... h (l <=...<= k1 <= ... k2 <= ... <= h)
 
In this sequence: ( k1-l ) + ( h-k2 ) = ( h-l ) - ( k2-k1 ) <= ( h-l ); 
 
So (h - l) will be the maximum profit in this days.
 
Another case:
 
L1 ...d1... H1 K2 ...k... K3 L2 ...d2... H2 (L1 <=... H1 > K2 >=...k >=... K3 > L2 <=... H2 )
 
K2 ... K3 is not exist increase sequence.
 
then for any k in that position,
 
( k-d1 ) + ( d2-k ) <= ( K2-L1 ) + ( H2-K3 ) < ( H1-L1 ) + ( H2-L2 ) 
 
In my code, variant "start" is the start of every no decrease sequence.*/

A little Adjustment.

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

Best Time to Buy and Sell Stock III

（onlineJudge: https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/）

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

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思想：动态规划。 记录下从各位置（含）开始之前的最大利润和此时开始到最后的最大利润。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        vector<int> preProfile(prices.size()+2, 0), postProfile(prices.size()+2, 0);
 
        int minPrice = 0x7fffffff;
        for(size_t i = 1; i <= prices.size(); ++i) {
            minPrice = min(minPrice, prices[i-1]);
            preProfile[i] = max(prices[i-1] - minPrice, preProfile[i-1]);
        }
 
        int maxPrice = 0;
        for(int i = prices.size(); i >= 1; --i) {
            maxPrice = max(maxPrice, prices[i-1]);
            postProfile[i] = max(maxPrice - prices[i-1], postProfile[i+1]);
        }
 
        int maxProfile = 0;
        for(size_t i = 1; i <= prices.size(); ++i)
            maxProfile = max(maxProfile, preProfile[i] + postProfile[i]);
        return maxProfile;
    }
};