121. Best Time to Buy and Sell Stock ------E

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

 

Example 1:

Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

 

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

A:
 ------------------3 rd pass -----------------
--就是很简单的一遍过即可---------不知道为什么，当时非要那么麻烦-----------------------------------------

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

就是CLRS上的，divide and conquer 问题。
第一遍做的，有点儿问题。（ 虽然也pass了）
例如： 上来直接设  leftMax =0  是不对的。  有可能很小。  因为，对于cross的， 就是假定了， 左右，都必须含有。因此，需要先设为： 负无穷 （ CLRS上也是这样的）

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices!= null && prices.length<=1){
            return 0;
        }
        int[] change = new int[prices.length];
        for(int i =0; i<prices.length-1;i++){
            change[i] = prices[i+1] - prices[i];
        }
        return maxChange(change, 0 ,change.length-1);
    }
    //divide and conquers problem.
    public int maxChange(int[] change, int start, int end){
        if(start==end){
            return change[start];
        }
        else{
            int middle = (int)((start+end)/2);
            int leftMax = maxChange(change, start,middle);
            int rightMax = maxChange(change, middle+1, end);
            int crossMax = maxCross(change,start,middle,end);
            int maxProfit = (leftMax>rightMax)?leftMax:rightMax;
            if(crossMax > maxProfit){
                maxProfit = crossMax;
            }
            return maxProfit;
        }
        
    }
    private int maxCross(int[] change, int start, int middle, int end){
        //get max left part
        int leftMax = 0;
        int sum=0;
        for(int i =middle; i>=start; i--){
            sum += change[i];
            if(sum>leftMax){
                leftMax = sum;
            }
        }
        int rightMax = 0;
        sum = 0;
        for(int i =middle+1; i<=end; i++){
            sum +=change[i];
            if(sum>rightMax){
                rightMax=sum;
            }
        }
        return leftMax + rightMax;
    }
}