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  • LeetCode

    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.
    class Solution {
        class Athlete {
            int index;
            int score;
            public Athlete(int index, int score) {
                this.index = index;
                this.score = score;
            }
        }
        public String[] findRelativeRanks(int[] nums) {
            if (nums == null || nums.length <= 0)
                return new String[0];
            String[] ret = new String[nums.length];
            Athlete[] athletes = new Athlete[nums.length];
            for (int i=0; i<nums.length; i++) {
                athletes[i] = new Athlete(i, nums[i]);
            }
            Arrays.sort(athletes, new Comparator<Athlete>() {
                @Override
                public int compare(Athlete o1, Athlete o2) {
                    return o2.score - o1.score;
                }
            });
            for (int i=0; i<athletes.length; i++) {
                Athlete a = athletes[i];
                if (i == 0)
                    ret[a.index] = "Gold Medal";
                else if (i == 1)
                    ret[a.index] = "Silver Medal";
                else if (i == 2)
                    ret[a.index] = "Bronze Medal";
                else 
                    ret[a.index] = String.valueOf(i+1);
            }
            return ret;
        }
    }
     
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  • 原文地址:https://www.cnblogs.com/wxisme/p/9487350.html
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