121.Best time to Buy and Sell Stock

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Dynamic Programming, Easy

Question

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

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

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Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

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Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Eg：

[7,1,5,3,6,4], difference between adjacent elements is [-6, 5, -2, 3, -2]

The problem is changed to find Largest continuous substring

My Answer

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class Solution {
    public int maxProfit(int[] prices) {
        int diff=0, temp, bigger, res=0;
        for(int i=0;i<prices.length-1;i++){
            temp = prices[i];
            bigger = prices[i+1];
            for(int j=i+2;j<prices.length;j++)            
                if(prices[j]>bigger)
                    bigger = prices[j];
            diff = bigger-temp;
            if(res<diff)
                res=diff;
        }
        return res;
    }
}

Running time: $O(n^2)$

With less time:

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class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length<=1)
            return 0;   
        int temp[] = new int[prices.length-1];
        for(int i=0;i<temp.length;i++)
            temp[i] = prices[i+1]-prices[i];
        
        int f[] = new int[temp.length];
        int res = temp[0];
        f[0] = temp[0];
        for(int i=1;i<temp.length;i++){
            if(f[i-1]<0)
                f[i] = temp[i];
            else
                f[i] = f[i-1] + temp[i];
            res = Math.max(res, f[i]);
        }
        return Math.max(res, 0);
    }
}

Running time: O(n) 4ms