Pascal's Triangle

Given a non-negative integer numRows, generate the first _numRows _of Pascal's triangle.


In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:

Input:
 5

Output:

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

Analysis

DP的初始条件也就是第一行[1], 第二行[1, 1]

每一层之间的递推关系很直观:

row[i] = lastRow[i] + lastRow[j - 1]

因此只需要按此写出代码即可。

Solution

Dynamic Programming O(n^2) time (0ms 100% AC)

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();

        for (int i = 0; i < numRows; i++) {
            List<Integer> row = new ArrayList<Integer>();
            List<Integer> lastRow = i > 0 ? res.get(i - 1) : new ArrayList<Integer>();
            for (int j = 0; j <= i; j++) {
                if (j == 0 || j == i) {
                    row.add(1);
                } else {
                    row.add(lastRow.get(j) + lastRow.get(j - 1));
                }
            }
            res.add(row);
        }
        return res;
    }
}

Another Implementation

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();

        // 1st base case
        if (numRows == 0) return res;

        // 2nd base case
        res.add(new ArrayList<>());
        res.get(0).add(1);

        for (int i = 1; i < numRows; i++) {
            List<Integer> row = new ArrayList<Integer>();
            List<Integer> lastRow = res.get(i - 1);
            for (int j = 0; j <= i; j++) {
                if (j == 0 || j == i) {
                    row.add(1);
                } else {
                    row.add(lastRow.get(j) + lastRow.get(j - 1));
                }
            }
            res.add(row);
        }
        return res;
    }
}

DP (LeetCode official answer)

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> triangle = new ArrayList<List<Integer>>();

        // First base case; if user requests zero rows, they get zero rows.
        if (numRows == 0) {
            return triangle;
        }

        // Second base case; first row is always [1].
        triangle.add(new ArrayList<>());
        triangle.get(0).add(1);

        for (int rowNum = 1; rowNum < numRows; rowNum++) {
            List<Integer> row = new ArrayList<>();
            List<Integer> prevRow = triangle.get(rowNum-1);

            // The first row element is always 1.
            row.add(1);

            // Each triangle element (other than the first and last of each row)
            // is equal to the sum of the elements above-and-to-the-left and
            // above-and-to-the-right.
            for (int j = 1; j < rowNum; j++) {
                row.add(prevRow.get(j-1) + prevRow.get(j));
            }

            // The last row element is always 1.
            row.add(1);

            triangle.add(row);
        }

        return triangle;
    }
}

DP - 2D array - O(n^2) time, O(n^2) space

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> result = new ArrayList<>();

        if (numRows == 0) {
            return result;
        }

        int[][] dp = new int[numRows][numRows];
        List<Integer> row;
        for (int i = 0; i < numRows; i++) {
            row = new ArrayList<Integer>();
            dp[i][0] = 1;
            row.add(dp[i][0]);
            for (int j = 1; j <= i; j++) {
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                row.add(dp[i][j]);
            }
            result.add(row);
        }
        return result;
    }
}

Reference

https://leetcode.com/problems/pascals-triangle/solution/

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