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 dayi.

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.

這是一個動態規劃問題，maxpro、minprice是在這一個陣列中不斷更新變化的資料。maxpro=pirce[i]-minprice即可。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int maxPro = 0;
        int minPrice = INT_MAX;
        for(int i = 0; i < prices.size(); i++){
            minPrice = min(minPrice, prices[i]);
            maxPro = max(maxPro, prices[i] - minPrice);
        }
        return maxPro;
    }
};
 

House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonightwithout alerting the police.

對於這類求極值的問題考慮動態規劃Dynamic Programming來解，在一個一位陣列中，其中f[i]表示到i位置時不相鄰數能形成的最大和，當我們到達一個arr[i]時我們考慮該房間和arr[i-1]的前一個房間之和（要求不相鄰）與f[i]進行比較而更新f[i]，我們可以得到遞推公式f[i] = max(num[i] + dp[i - 2], dp[i - 1]), 由此看出我們需要初始化f[0]和f[1]，其中f[0]即為arr[0]，f[1]此時應該為max(arr[0], arr[1])

class Solution {
public:
    int rob(vector<int>& nums) {
        const int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        if (n == 2) return max(nums[0], nums[1]);
        vector<int> f(n, 0);
        f[0] = nums[0];
        f[1] = max(nums[0], nums[1]);
        for (int i = 2; i < n; ++i)
            f[i] = max(f[i-2] + nums[i], f[i-1]);
        return f[n-1];
    }
};