2022. Convert 1D Array Into 2D Array

题目

You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original.

The elements from indices 0 to n – 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n – 1 (inclusive) should form the second row of the constructed 2D array, and so on.

Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.

Example 1:

Input: original = [1,2,3,4], m = 2, n = 2
Output: [[1,2],[3,4]]
Explanation: The constructed 2D array should contain 2 rows and 2 columns.
The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.


Example 2:

Input: original = [1,2,3], m = 1, n = 3
Output: [[1,2,3]]
Explanation: The constructed 2D array should contain 1 row and 3 columns.
Put all three elements in original into the first row of the constructed 2D array.


Example 3:

Input: original = [1,2], m = 1, n = 1
Output: []
Explanation: There are 2 elements in original.
It is impossible to fit 2 elements in a 1×1 2D array, so return an empty 2D array.
 

Constraints:

1 <= original.length <= 5 * 104
1 <= original[i] <= 105
1 <= m, n <= 4 * 104

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/convert-1d-array-into-2d-array
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

我的题解

class Solution {
    public int[][] construct2DArray(int[] original, int m, int n) {
        if(m*n != original.length){
            return new int[][]{};
        }
        int[][] a2d = new int[m][n];
        int r = 0,c = 0;
        for(int i=0;i<original.length;i++){
            a2d[r][c++] = original[i];
            if((i+1)%n==0){
                r++;
                c = 0;
            }
        }
        return a2d;
    }
}

基本与官方题解差不多,官方题解在循环时i以n的倍数递增,然后使用数组拷贝函数进行拷贝。数组索引i从0开始, i/n 对应 i 所在一维数组在二维数组中的索引(即行数)

执行用时:4 ms, 在所有 Java 提交中击败了52.55%的用户
内存消耗:49.2 MB, 在所有 Java 提交中击败了5.09%的用户

方法一: 参考官方解法

    public int[][] construct2DArray(int[] original, int m, int n) {
        if(m*n != original.length){
            return new int[0][0];
        }
        int[][] a2d = new int[m][n];
        for(int i=0; i<m; i++){
            System.arraycopy(original, n*i, a2d[i], 0, n);
        }
        return a2d;
    }

方法二:宫水三叶的解法

    public int[][] construct2DArray(int[] original, int m, int n) {
        if(m*n != original.length){
            return new int[0][0];
        }
        int[][] a2d = new int[m][n];
        int idx = 0;
        for(int i=0; i<m; i++){
            for(int j=0;j<n;j++){
                a2d[i][j] = original[idx++];
            }
        }
        return a2d;
    }

官方题解

class Solution {
    public int[][] construct2DArray(int[] original, int m, int n) {
        if (original.length != m * n) {
            return new int[0][];
        }
        int[][] ans = new int[m][n];
        for (int i = 0; i < original.length; i += n) {
            System.arraycopy(original, i, ans[i / n], 0, n);
        }
        return ans;
    }
}

作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/convert-1d-array-into-2d-array/solution/jiang-yi-wei-shu-zu-zhuan-bian-cheng-er-zt47o/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

Leave a Reply

Your email address will not be published. Required fields are marked *