LeetCode解题报告（505）-- 2099. Find Subsequence of Length K With the Largest Sum

Problem

You are given an integer array nums and an integer k. You want to find a subsequence of nums of length k that has the largest sum.

Return *any such subsequence as an integer array of length* k.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [2,1,3,3], k = 2
Output: [3,3]
Explanation:
The subsequence has the largest sum of 3 + 3 = 6.

Example 2:

Input: nums = [-1,-2,3,4], k = 3
Output: [-1,3,4]
Explanation: 
The subsequence has the largest sum of -1 + 3 + 4 = 6.

Example 3:

Input: nums = [3,4,3,3], k = 2
Output: [3,4]
Explanation:
The subsequence has the largest sum of 3 + 4 = 7. 
Another possible subsequence is [4, 3].

Constraints:

• 1 <= nums.length <= 1000
• -105 <= nums[i] <= 105
• 1 <= k <= nums.length

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Analysis

  这道题目给出一个数组，要求求出拥有最大和的长度为k的子序列。因为是子序列，不是子数组，所以不考虑sliding window。实际上就是取前k大的数，但是题目要求返回的是数组，而且元素之间的相对位置要保持不变。所以我们就不能只用一个数组来做了。首先copy一个数组出来，然后排序，取前k大的数放到unordered_map中，然后再遍历一遍原数组，只有在unordered_map中出现过的才放到结果中，注意动态更新，这样就能按照原来的相对顺序得到答案。

Solution

  无。

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Code

class Solution {
public:
    vector<int> maxSubsequence(vector<int>& nums, int k) {
        vector<int> temp = nums;
        sort(nums.rbegin(), nums.rend());
        unordered_map<int, int> m;
        for (int i = 0; i < k; ++i) {
            ++m[nums[i]];
        }
        
        vector<int> result;
        for (int num: temp) {
            if (m[num]-- > 0) {
                result.push_back(num);
            }
        }
        return result;
    }
};

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Summary

  这道题目虽然不难，但是如果没有做过的话，一下子还是有点没想过来。这道题目的分享到这里，感谢你的支持！