动态规划两题

动态规划两题

算法概论第十三周

123. Best Time to Buy and Sell Stock III

123. Best Time to Buy and Sell Stock III

题目描述

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 at most two transactions.

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

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 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.

思路分析

• 这一题如果只允许进行一次交易，比较简单
  • 设dp[i]是第i天最大收益，min为历史最低价，maxprofit为历史最大收益

int maxProfit = 0;
int minNum = prices[0];
dp[0] = 0;
for(int d = 1; d < dayNum; d++){
    int diff = prices[d] - minNum;
    if(diff < 0){
        minNum = prices[d];
    }
    else{
        maxProfit = max(maxProfit, diff);
    }
    dp[d] = maxProfit;
}

• 但如果考虑两次交易也没关系，枚举每一天作为分割点，改天前半部分和后半部分分别求对应的最大收益即可

int res = 0;
for(int d = 0; d < dayNum; d++){
    res = max(res, maxBeforeDay[d] + maxAfterDay[d]);
}
return res;

实现代码

const int MAXN = 100000;
int maxBeforeDay[MAXN] = {0};
int maxAfterDay[MAXN] = {0};
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int dayNum = prices.size();
        if(dayNum <= 1)
        return 0;

        int maxProfit = 0;
        int minNum = prices[0];
        maxBeforeDay[0] = 0;
        for(int d = 1; d < dayNum; d++){
            int diff = prices[d] - minNum;
            if(diff < 0){
                minNum = prices[d];
            }
            else{
                maxProfit = max(maxProfit, diff);
            }
            maxBeforeDay[d] = maxProfit;
        }

        int maxNum = prices[dayNum - 1];
        maxProfit = 0;
        maxAfterDay[dayNum - 1] = 0;
        for(int d = dayNum - 2; d>=0; d--){
            int diff = maxNum - prices[d];
            if(diff < 0){
                maxNum = prices[d];
            }
            else{
                maxProfit = max(maxProfit, diff);
            }
            maxAfterDay[d] = maxProfit;
        }
        int res = 0;
        for(int d = 0; d < dayNum; d++){
            res = max(res, maxBeforeDay[d] + maxAfterDay[d]);
        }
        return res;
    }
};

120. Triangle

120. Triangle —— 题目链接

题目描述

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

思路分析

• 每个节点只需要考虑其两个子节点向上路径的最短值
• 太简单了，不过瘾，我再做一题

实现代码

class Solution {
public:
    
    int minimumTotal(vector<vector<int>>& triangle) {
        int dp[10000] = {};
        int height = triangle.size();
        if(height == 1){
            return triangle[0][0];
        }
        for(int i = 0; i < triangle[height - 1].size(); i++){
            dp[i] = triangle[height - 1][i];
        }
        for(int i = height - 2; i>=0; i--){
            int tdp[10000] = {};
            for(int j = triangle[i].size() - 1; j>=0; j--){
                tdp[j] = triangle[i][j] + min(dp[j], dp[j + 1]);
            }
            for(int k = 0; k < triangle[i].size(); k++){
                dp[k] = tdp[k];
            }
        }
        return dp[0];
    }
};