【leetcode】123. Best Time to Buy and Sell Stock III

@requires_authorization
@author johnsondu
@create_time 2015.7.22 19:04
@url [Best Time to Buy and Sell Stock III](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/)
/************************
 * @description: dynamic programming.
 *    从前后分别求出当前所能够得到的最大值，最后相加而得
 *    详细操作是，设置一个最小值，然后不断更新最小值。然后不断更新最大值。

前后反向累加求得最大值
 * @time_complexity:  O(n)
 * @space_complexity: O(n)
 ************************/
 class Solution {
public:
    int maxProfit(vector<int>& prices) {
        const int len = prices.size();
        if(len < 2) return 0;
        int maxFromHead[len];
        maxFromHead[0] = 0;
        int minprice = prices[0], maxprofit = 0;
        for(int i = 1; i < len; i ++) {
            minprice = min(prices[i-1], minprice);
            if(maxprofit < prices[i] - minprice)
                maxprofit = prices[i] - minprice;
            maxFromHead[i] = maxprofit;
        }
        int maxprice = prices[len-1];
        int res = maxFromHead[len-1];
        maxprofit = 0;
        for(int i = len-2; i >= 0; i --) {
            maxprice = max(maxprice, prices[i+1]);
            if(maxprofit < maxprice - prices[i])
                maxprofit = maxprice - prices[i];
            if(res < maxFromHead[i] + maxprofit)
                res = maxFromHead[i] + maxprofit;
        }
        return res;
    }
};

@requires_authorization
@author johnsondu
@create_time 2015.7.22 19:04
@url [Best Time to Buy and Sell Stock III](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/)
/************************
 * @description: dynamic programming.
 *    超时做法：从1開始。分成前后两段，最后求得前后两段的最大值
 * @time_complexity:  O(n^2)
 * @space_complexity: O(n)
 ************************/
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int p_size = prices.size();
        if(p_size < 2) return 0;
        if(p_size < 3) {
            return prices[1] - prices[0] > 0 ?

prices[1] - prices[0]: 0;
        }

        int ans = 0;
        for(int i = 1; i < p_size; i ++) {
            vector<int> dp1, dp2;
            for(int j = 1; j <= i; j ++) {
                dp1.push_back(prices[i] - prices[i-1]);
            }
            for(int j = i + 1; j < p_size; j ++) {
                dp2.push_back(prices[j] - prices[j-1]);
            }

            int dp1_size = dp1.size();
            int ans1 = 0;
            int cur = 0;
            for(int j = 0; j < dp1_size; j ++) {
                cur = cur + dp1[j];
                if(cur < 0) {
                    cur = 0;
                    continue;
                }
                if(cur > ans1) ans1 = cur;
            }

            int dp2_size = dp2.size();
            int ans2 = 0;
            cur = 0;
            for(int j = 0; j < dp2_size; j ++) {
                cur = cur + dp2[j];
                if(cur < 0) {
                    cur = 0;
                    continue;
                }
                if(cur > ans2) ans2 = cur;
            }
            ans = max(ans, ans1 + ans2);
        }
        return ans;
    }
};