leetcode 121. 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 day i.

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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

本质上都是计数器，在某些时候reset计数。

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices: return 0
        low = high = prices[0]
        ans = 0
        for i in xrange(1, len(prices)):
            if prices[i] > high: # 看到高价就贪心
                high = prices[i]
                ans = max(ans, high-low)
            elif prices[i] < low:
                low = high = prices[i] # 遇到更低价格就reset
        return ans

或者是：

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;
}

另外的解法：

The logic to solve this problem is same as “max subarray problem” using Kadane's Algorithm. Since no body has mentioned this so far, I thought it’s a good thing for everybody to know.

All the straight forward solution should work, but if the interviewer twists the question slightly by giving the difference array of prices, Ex: for {1, 7, 4, 11}, if he gives {0, 6, -3, 7}, you might end up being confused.

Here, the logic is to calculate the difference (maxCur += prices[i] - prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below 0, reset it to zero.

    public int maxProfit(int[] prices) {
        int maxCur = 0, maxSoFar = 0;
        for(int i = 1; i < prices.length; i++) {
            maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);
            maxSoFar = Math.max(maxCur, maxSoFar);
        }
        return maxSoFar;
    }

*maxCur = current maximum value

*maxSoFar = maximum value found so far

举例（遇到拐点）：股票价格为【1,3,2,7】，则3-1=2，2-3=-1，-1+2=1（非常巧妙），7-2=5，5+1=6，所以答案是6.