LeetCode----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 mosttwo transactions.

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

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思路：循环利用解决best-time-to-buy-and-sell-stock-i（一次交易的最大收益）的方法，将数组分为front和back两部分并分别求其中的一个最大收益作和。

           每次更新保留较大收益。

代码实现：

C/C++ (clang++ 3.3)

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class Solution {

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public:

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    int maxProfit(vector<int> &prices) {

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        if (prices.size()<=1)

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            return 0;

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        int mx = 0;//mx记录当前最大收益

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        vector<int>::iterator iter;

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        for (iter=prices.begin(); iter!=prices.end(); ++iter){

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            //将数组分为前后两部分并分别计算最大收益

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            vector<int> front(prices.begin(), iter+1);

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            vector<int> back(iter+1, prices.end());

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            int a = maxP(front);

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            int b = maxP(back);

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            mx =max(a+b, mx); //更新当前最大收益

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        }

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        return mx;

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    }

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    int maxP(vector<int> &price){//处理一次交易最大收益

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        if (price.size() <= 1)

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            return 0;

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        int mx = 0;

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        for (int i=0; i<price.size()-1; ++i){

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            for (int j=i+1; j<price.size(); ++j){

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                mx = max(mx, price[j]-price[i]);

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            }

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        }

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        return mx;

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    }

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};

您的代码已保存

网上 波峰波谷 解答：

classSolution {

public:

    intmaxProfit(vector<int> &prices) {

        if(prices.size() < 2)

            return0;

        constint n = prices.size();

        vector<int> f(n, 0);

        vector<int> g(n, 0);

        for(inti = 1, valley = prices[0]; i < n; ++i) {

            valley = min(valley, prices[i]);

            f[i] = max(f[i - 1], prices[i] - valley);

        }

        for(inti = n - 2, peak = prices[n - 1]; i >= 0; --i) {

            peak = max(peak, prices[i]);

            g[i] = max(g[i], peak - prices[i]);

        }

        intmax_profit = 0;

        for(inti = 0; i < n; ++i)

            max_profit = max(max_profit, g[i] + f[i]);

         

        returnmax_profit;

    }

};

------------------------------------------------------------

------------------------------------------------------------

class Solution {

public:

    int maxProfit(vector<int> &prices) {

        //You may complete at most two transactions.(you must sell the stock before you buy again)

        //方式一：爆搜 (630ms 8552k)

        /*

        int len=prices.size();

        int maxPro=0;

        for(int i=0;i<len;i++){ //以i作为两次交易的分界线

            int max1=0,max2=0;

            for(int j=0;j<=i;j++){

                for(int k=j+1;k<=i;k++){

                    int val = prices[k]-prices[j];

                    if(val>max1){

                        max1=val;

                    }

                }

            }

            for(int j=i;j<len;j++){

                for(int k=j+1;k<len;k++){

                    int val = prices[k]-prices[j];

                    if(val>max2){

                        max2=val;

                    }

                }

            }

            if(max1+max2>maxPro)maxPro = max1+max2;

        }

        return maxPro;

        */

         

         

        //方式二：波峰波谷检测法，可以减小不必要的搜索空间,相比于第一种效率提升了不少    (180ms 8568k)

        int len = prices.size();

        if(len<=1)return 0;

        vector<int> flag(len,0);

        for(int i=1;i<len-1;i++){

            if(prices[i]>prices[i-1]&&prices[i]>=prices[i+1])flag[i]=1;  //标记为波峰(比较时注意相等情况)

            if(prices[i]<prices[i-1]&&prices[i]<=prices[i+1])flag[i]=-1;  //标记为波谷

        }

        if(prices[0]<=prices[1])flag[0]=-1;

        if(prices[len-1]>prices[len-2])flag[len-1]=1;

        //搜索

        int maxPro=0;

        for(int k=0;k<len;k++){

            int max1=0,max2=0;

            for(int i=0;i<=k;i++){

                if(flag[i]==-1){

                    for(int j=i+1;j<=k;j++){

                        if(flag[j]==1){

                            int val = prices[j]-prices[i];

                            if(val>max1)max1 = val;

                        }

                    }

                }

            }

            for(int i=k;i<len;i++){

                if(flag[i]==-1){

                    for(int j=i+1;j<len;j++){

                        if(flag[j]==1){

                            int val = prices[j]-prices[i];

                            if(val>max2)max2 = val;

                        }

                    }

                }

            }

            int val = max1+max2;

            if(val>maxPro)maxPro = val;

        }

        return maxPro;

    }

};