506. Relative Ranks

Given scores of N athletes, find their relative ranks and the people with the top three highest scores, who will be awarded medals: "Gold Medal", "Silver Medal" and "Bronze Medal".

Example 1:

Input: [5, 4, 3, 2, 1]
Output: ["Gold Medal", "Silver Medal", "Bronze Medal", "4", "5"]
Explanation: The first three athletes got the top three highest scores, so they got "Gold Medal", "Silver Medal" and "Bronze Medal". 
For the left two athletes, you just need to output their relative ranks according to their scores.

思路：对于给予的得分情况，找出前三名并给予相应的称号，其余以数字作为其名词，利用优先队列进行记录每个元素的位置和元素值，优先队列的特点是top始终是所有元素中最大的那个。

class Solution {
public:
    vector<string> findRelativeRanks(vector<int>& nums) {
        vector<string> res(nums.size());
        priority_queue<pair<int,int>> temp;
        for(size_t i=0,len=nums.size();i<len;++i)
            temp.push({nums[i],i});
        int count=1;
        while(temp.size())
        {
            auto elem=temp.top();
            temp.pop();
            if(count==1)
                res[elem.second]="Gold Medal";
            else if(count==2)
                res[elem.second]="Silver Medal";
            else if(count==3)
                res[elem.second]="Bronze Medal";
            else
                res[elem.second]=to_string(count);
            ++count;
        }
        return res;
        
    }
};

第二种方法，可以用map来解决，将nums中的元素值作为map的index从而实现将nums排序，同时保存nums的index到map中

class Solution {
public:
    vector<string> findRelativeRanks(vector<int>& nums) {
        vector<string> res(nums.size());
        map<int,int>ma;
        for(int i=0;i!=nums.size();++i)
            ma[nums[i]]=i;
        int count=1;
        for(map<int,int>::reverse_iterator it=ma.rbegin();it!=ma.rend();++it,++count)
        {
            if(count==1)
                res[it->second]="Gold Medal";
            else if(count==2)
                res[it->second]="Silver Medal";
            else if(count==3)
                res[it->second]="Bronze Medal";
            else
                res[it->second]=to_string(count);
        }
        return res;
        
    }
};