Complete the function below. The function receives the number of rows and returns the generated Pascal's Triangle. Problem: Pascal's Triangle Given a non-negative integer numRows, generate the first numRows rows of Pascal's Triangle and return them as a 2D array triangle. Pascal's Triangle is defined as: Row i (0-indexed) contains i+1 elements. The first and last element of each row is 1. For other positions: triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j] for 0 < j < i. Input A single integer: numRows Output Print a 2D array (you may print each row as a list) representing the first numRows rows. Constraints 0 ≤ numRows ≤ 30 Example Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] Example Input 0 Output [] Function Description Complete solvePascalsTriangle. It has one parameter, int numRows. Return the first numRows rows of Pascal's Triangle as a 2D integer array.

Constraints

Use the limits and requirements stated in the prompt.

解法

逐行构造：第 0 行为 [1]，第 i 行除首尾的 1 外，row[j] = prev[j-1] + prev[j]。时间复杂度 O(numRows²)，空间复杂度 O(numRows²) 用于结果。

from typing import List

def solvePascalsTriangle(numRows: int) -> List[List[int]]:
    res = []
    for i in range(numRows):
        row = [1] * (i + 1)
        for j in range(1, i):
            row[j] = res[i - 1][j - 1] + res[i - 1][j]
        res.append(row)
    return res

class Solution {
    int[][] solvePascalsTriangle(int numRows) {
        int[][] res = new int[numRows][];
        for (int i = 0; i < numRows; i++) {
            res[i] = new int[i + 1];
            res[i][0] = res[i][i] = 1;
            for (int j = 1; j < i; j++) {
                res[i][j] = res[i - 1][j - 1] + res[i - 1][j];
            }
        }
        return res;
    }
}

class Solution {
public:
    vector<vector<int>> solvePascalsTriangle(int numRows) {
        vector<vector<int>> res(numRows);
        for (int i = 0; i < numRows; i++) {
            res[i].assign(i + 1, 1);
            for (int j = 1; j < i; j++) {
                res[i][j] = res[i - 1][j - 1] + res[i - 1][j];
            }
        }
        return res;
    }
};

Given an integer array nums and an integer k, return the k most frequent elements. To make the output deterministic, sort elements by frequency in descending order. If two elements have the same frequency, the smaller numeric value comes first. Function Description Complete solveTopKFrequentElements. It has the following parameters: int[] nums: the input array int k: the number of elements to return Return an int[] containing the top k elements in the deterministic order described above.

Constraints

1 ≤ nums.length ≤ 10⁵ -10⁴ ≤ nums[i] ≤ 10⁴ 1 ≤ k ≤ number of distinct elements in nums

Example 1

Input:

nums = [1,1,1,2,2,3]
k = 2

Output:

[1,2]

Explanation: 1 appears 3 times and 2 appears 2 times, so they are the two most frequent elements.

Example 2

Input:

nums = [4,5,2,1,3]
k = 5

Output:

[1,2,3,4,5]

Explanation: Every number appears once, so ties are resolved by smaller value first.