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.

Note:

1. N is a positive integer and won't exceed 10,000.
2. All the scores of athletes are guaranteed to be unique.

这道题给了我们一组分数，让我们求相对排名，前三名分别是金银铜牌，后面的就是名次数，不是一道难题，我们可以利用堆来排序，建立一个优先队列，把分数和其坐标位置放入队列中，会自动按其分数高低排序，然后我们从顶端开始一个一个取出数据，由于保存了其在原数组的位置，我们可以直接将其存到结果res中正确的位置，用一个变量cnt来记录名词，前三名给奖牌，后面就是名次数，参见代码如下：

解法一：

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

下面这种方法思路和上面一样，不过数据结构用的不同，这里利用map的自动排序的功能，不过map是升序排列的，所以我们遍历的时候就要从最后面开始遍历，最后一个是金牌，然后往前一次是银牌，铜牌，名次数等，参见代码如下：

解法二：

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

下面这种方法没用什么炫的数据结构，就是数组，建立一个坐标数组，不过排序的时候比较的不是坐标，而是该坐标位置上对应的数字，后面的处理方法和之前的并没有什么不同，参见代码如下：

解法三：

class Solution {
public:
    vector<string> findRelativeRanks(vector<int>& nums) {
        int n = nums.size();
        vector<int> idx(n);
        vector<string> res(n, "");
        for (int i = 0; i < n; ++i) idx[i] = i;
        sort(idx.begin(), idx.end(), [&](int a, int b){return nums[a] > nums[b];});
        for (int i = 0; i < n; ++i) {
            if (i == 0) res[idx[i]] = "Gold Medal";
            else if (i == 1) res[idx[i]] = "Silver Medal";
            else if (i == 2) res[idx[i]] = "Bronze Medal";
            else res[idx[i]] = to_string(i + 1);
        }
        return res;
    }
};

参考资料：

https://discuss.leetcode.com/topic/77912/c-easy-to-understand

https://discuss.leetcode.com/topic/77876/easy-java-solution-sorting

https://discuss.leetcode.com/topic/78244/simple-c-solution-using-a-map

https://discuss.leetcode.com/topic/77869/simple-sorting-o-n-log-n-solution

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