Best Time to Buy and Sell Stock

最新推荐文章于 2025-04-18 22:24:21 发布

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本文介绍了三种求解股票买卖最大利润的算法实现方法，包括分治法、一次遍历法和动态规划法。通过这些方法可以高效地计算出在给定价格序列中能够获得的最大利润。

I: Divide and Conquer

class Solution {
public:
    int max_Diff(vector<int> &prices, int start, int end, int *max, int *min) {
        if (start>=end) {
            *max=*min=prices[start];
            return 0;
        }
        int mid=(start+end)/2;
        int leftMax=0,leftMin=0,rightMax=0,rightMin=0;
        int leftDiff=max_Diff(prices, start, mid, &leftMax, &leftMin);
        int rightDiff=max_Diff(prices, mid+1, end, &rightMax, &rightMin);
        *max=(leftMax>rightMax)?leftMax:rightMax;
        *min=(leftMin<rightMin)?leftMin:rightMin;
        int maxDiff=rightMax-leftMin;
        maxDiff=(maxDiff>leftDiff)?maxDiff:leftDiff;
        maxDiff=(maxDiff>rightDiff)?maxDiff:rightDiff;
        return maxDiff;
    }
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (prices.size()<2) {
            return 0;
        }
        int min=0,max=0;
        return max_Diff(prices,0,prices.size()-1,&max,&min);
        
    }
};

遍历一遍，每次用当前的减之前最小的是当前的最大利润，更新最大利润。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n=prices.size();
        if (!n) return 0;
        int max=0, min=prices[0];
        for (int i=1; i<n; i++) {
            if (prices[i]-min>max) {
                max = prices[i]-min;
            }
            if (prices[i]<min) {
                min=prices[i];
            }
        }
        return max;
    }
};

II: 所有差的和

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n=prices.size();
        if (n<2) return 0;
        int res=0;
        for (int i=1; i<n; i++) {
            if (prices[i]>prices[i-1]) {
                res += prices[i]-prices[i-1];
            }
        }
        return res;
    }
};

III：DP

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(prices.size()<2) {
            return 0;
        }
        int res=0;
        vector<int> v1(prices.size());
        vector<int> v2(prices.size());
        int min=0;
        for (int i=1; i<prices.size(); i++) {
            if(prices[i]<prices[min]) {
                min=i;
            }
            v1[i]=v1[i-1]>(prices[i]-prices[min])?v1[i-1]:(prices[i]-prices[min]);
        }
        int max=prices.size()-1;
        for (int i=prices.size()-2; i>0; i--) {
            if(prices[i]>prices[max]) {
                max=i;
            }
            v2[i]=v2[i+1]>(prices[max]-prices[i])?v2[i+1]:(prices[max]-prices[i]);
        }
        for (int i=0; i<prices.size(); i++) {
            res=res>(v1[i]+v2[i])?res:(v1[i]+v2[i]);
        }
        return res;
    }
};

class Solution {
public:
    int maxProfit(vector &prices) {
        int max=0, n=prices.size();
        if (n==0) return 0;
        int low=INT_MAX, high=prices[n-1];
        vector profit(n,0);
        for (int i=0; i<n; i++) {
            if (prices[i]max) max=cur;
            profit[i]=max;
        }
        for (int i=n-2; i>0; i--) {
            if (prices[i]>high) high=prices[i];
            int cur=profit[i-1]+high-prices[i];
            if (cur>max) max=cur;
        }
        return max;
    }
};