leetcode 566 Reshape the Matrix-优快云博客

本文介绍了一种将二维数组重塑为特定尺寸矩阵的方法，并提供了一个Java实现示例。该算法通过遍历原始矩阵，按顺序填充目标矩阵，保持原有元素的顺序。

题目详情

You're given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.
The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.
If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
题目要求输入一个二维数组代表的矩阵、期望矩阵的长、期望矩阵的宽，我们根据长宽对它进行“重塑”。输入矩阵和要求的输出矩阵的矩阵元素数应该是相等的，否则就是不合法的输入，直接返回输入矩阵即可。

Example 1:
Input: nums = [[1,2],[3,4]]
r= 1, c = 4
Output:
[[1,2,3,4]]

思路

我的想法比较简单，就是简单的遍历输入数组，同时把值赋给结果数组的相应位置即可。

解法

    public int[][] matrixReshape(int[][] nums, int r, int c) {
        if(nums == null || nums.length == 0 || r*c != nums.length*nums[0].length) return nums;
        int heightIndex = 0;
        int widthIndex = 0;
        //保存结果数组
        int[][] res = new int[r][c];
        for(int i =0;i<nums.length;i++){
            for(int j=0;j<nums[0].length;j++){
                res[heightIndex][widthIndex] = nums[i][j];
                widthIndex ++;
                if(widthIndex == c){
                    heightIndex ++;
                    widthIndex = 0;
                }
            }
        }
        return res;
    }