826. Most Profit Assigning Work

这一题做出来了，但是不必要的步骤比较多，也就是不够简洁。
自己的做法：
按照利润由高到低排序，如果某一些排在后面的难度却更高，就删掉它们。

class Solution {
public:
    int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
        int workSz = profit.size();
        if (workSz == 0)
            return 0;
        vector<pair<int, int>> works;
        for (int i = 0; i < workSz; ++i)
            works.push_back(make_pair(profit[i], difficulty[i]));
        sort(works.begin(), works.end(), sortHelper);
        int i = 1;
        int nowDif = works[0].second;
        for (int j = 1; j < workSz; ++j) {
            if (works[j].second < nowDif) {
                nowDif = works[j].second;
                works[i++] = works[j];
            }
        }
        //works.erase(works.begin()+i, works.end());必须删除范围
        int endIdx = i;
        sort(worker.begin(), worker.end(), greater<int>());
        int maxProfit = 0;
        i = 0;
        int j = 0;
        while (i < endIdx && j < worker.size()) {
            while (i < endIdx && works[i].second > worker[j])
                ++i;
            if (i == endIdx)
                break;
            maxProfit += works[i].first;
            ++j;
        }
        return maxProfit;
    }
private:
    static bool sortHelper(pair<int, int>& p1, pair<int, int>& p2) {
        if (p1.first > p2.first)
            return true;
        else if (p1.first < p2.first)
            return false;
        else
            return p1.second < p2.second;
    }
};

方法二：
更加简洁的。
直接sort，没有删除这个步骤。如果有那种“难度更大，收益更低”的工作，可以直接略过，不必要处理。

class Solution {
public:
    int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
        vector<pair<int, int>> jobs;
        for (int i = 0; i < profit.size(); ++i)
            jobs.push_back(make_pair(profit[i], difficulty[i]));
        sort(jobs.begin(), jobs.end());
        sort(worker.begin(), worker.end());
        int i = jobs.size()-1;
        int j = worker.size()-1;
        int maxProfit = 0;
        while (i >= 0 && j >= 0) {
            while (i >= 0 && jobs[i].second > worker[j])
                --i;
            if (i < 0)
                break;
            maxProfit += jobs[i].first;
            --j;
        }
        return maxProfit;
    }
};

这个就简洁的多。这个代码量才是正常的。上面的有点多了