leetcode983"最低票价问题"解法分析

自己解法一：

class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        int len = days.size();
        if (len == 0) {
            return 0;
        }
        vector<int> dp(len);
        dp[0] =
            *min_element(costs.begin(), costs.end());  // costs[0]; // !!!!天啦
        for (int i = 0; i < len - 1; i++) {
            vector<int> tmp;
            int k1 = -1;
            int k2 = -1;
            int k3 = -1;
            for (int j = i; j >= 0; j--) {
                if (days[i + 1] - days[j] + 1 > 1) {
                    k1 = j;
                }
            }
            if (k1 >= 0) { 
                tmp.push_back(dp[i] + costs[0]);
            } else {
                tmp.push_back(costs[0]);
            }
            for (int j = i; j >= 0; j--) {
                if (days[i + 1] - days[j] + 1 > 7) {
                    k2 = j;
                    break;
                }
            }
            if (k2 >= 0) {
                tmp.push_back(dp[k2] + costs[1]);
            } else {
                tmp.push_back(costs[1]);
            }
            for (int j = i; j >= 0; j--) {
                if (days[i + 1] - days[j] + 1 > 30) {
                    k3 = j;
                    break;
                }
            }
            if (k3 >= 0) {
                tmp.push_back(dp[k3] + costs[2]);
            } else {
                tmp.push_back(costs[2]);
            }
            dp[i + 1] = *min_element(tmp.begin(), tmp.end());
            cout << "i: " << i + 1 << " costs: " << dp[i + 1] << endl;
        }
        return dp[len - 1];
    }
};

自己解法二：回溯超时

class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        int len = days.size();
        if (len == 0) {
            return 0;
        }
        stack<vector<int>> s;
        stack<pair<int, int>> path;
        s.push({-1, -1, 0});
        path.push({-1, -1});
        vector<int> prices;
        vector<vector<int>> ans;
        while (s.empty() == false) {
            auto p = s.top();
            s.pop();
            path.push({p[0], p[1]});
            int l = p[0];
            int r = p[1];
            prices.push_back(p[2]);
            if (r == len - 1) {
                ans.push_back(prices);
                for (auto p : prices) {
                    cout << p << " ";
                }
                cout << endl;
                while (s.empty() == false && ((path.top().second + 1) != s.top().at(0))) {
                    path.pop();
                    prices.pop_back();
                }
            }
            for (int j = r + 1; j < len; j++) {
                if (days[j] - days[r + 1] + 1 >= 1) {
                    s.push({r + 1, j, costs[0]});
                    break;
                }
            }
            for (int j = r + 1; j < len; j++) {
                if (days[j] - days[r + 1] + 1 > 7) {
                    s.push({r + 1, j - 1, costs[1]});
                    break;
                } else if (j == len - 1) {
                    s.push({r + 1, len - 1, costs[1]});
                }
            }
            for (int j = r + 1; j < len; j++) {
                if (days[j] - days[r + 1] + 1 > 30) {
                    s.push({r + 1, j - 1, costs[2]});
                    break;
                } else if (j == len - 1) {
                    s.push({r + 1, len - 1, costs[2]});
                }
            } 
        }
        int minPrice = INT_MAX;
        for (auto price : ans) {
            int sum = accumulate(price.begin(), price.end(), 0);
            minPrice = (minPrice < sum) ? minPrice : sum;
        }
        return minPrice;
    }
};

力友解法三：

class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
	int sz = days.size();
	vector<int> dp(sz+1, 0);
	dp[1] = *min_element(costs.begin(), costs.end());
	int last7, last30;
	for (int i = 2; i<=sz; i++) {
		last7 = upper_bound(days.begin(), days.begin() + i, days[i-1] - 7) - days.begin();
		last30 = upper_bound(days.begin(), days.begin() + i, days[i-1] - 30) - days.begin();
		dp[i] = min(costs[0] + dp[i - 1], min(costs[1] + dp[last7], costs[2] + dp[last30]));
	}
	return dp[sz];
	}
};