Leetcode @ Max job profit in job scheduling

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime , endTime and profit arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

Constraints:

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
• 1 <= startTime[i] < endTime[i] <= 10^9
• 1 <= profit[i] <= 10^4

看到这道题感觉自觉会想到dp，贪心算法和 backtracking，它们之间有什么不同以及适合解决什么样的问题？可以总结一下。

问题1. startTime, endTime 是sorted么？如果不是的话，很可能我们需要先建一个data structure再自行sort. 

关键点：对于每一个开始时间，我们可以用map<int, int> dp来存放，map和unordered_map的差别是，map类似于BST，里面的数字是按照key升序排列，搜索是binary search O(logn)，插入一个节点rebalance也是O(logn),记得要初始化map[0] = 0，这是用来方便push in第一个初始工作的。

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
      int size = profit.size();
      vector<vector<int>> jobs;
      for (int i = 0; i<size; i++)
      {
          jobs.push_back({endTime[i], startTime[i], profit[i]});
      }
      sort(jobs.begin(), jobs.end());
      map<int, int> dp;
      dp.insert({0, 0});
      for (auto &job : jobs)
      {
          auto it = dp.upper_bound(job[1]);
          int cur = job[2] + prev(it)->second;
          if (cur > dp.rbegin()->second)
          {
              dp[job[0]] = cur;
          }
      }
        return dp.rbegin()->second;
    }
};