zoukankan      html  css  js  c++  java
  • 566. Reshape the Matrix

    In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.

    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.

    Example 1:

    Input: 
    nums = 
    [[1,2],
     [3,4]]
    r = 1, c = 4
    Output: 
    [[1,2,3,4]]
    Explanation:
    The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

    Example 2:

    Input: 
    nums = 
    [[1,2],
     [3,4]]
    r = 2, c = 4
    Output: 
    [[1,2],
     [3,4]]
    Explanation:
    There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

    Note:

    1. The height and width of the given matrix is in range [1, 100].
    2. The given r and c are all positive.

    大致意思就是把一个数组放到一个新数组,如果可以则返回新数组,如果不行则返回元数组。

    思路如下

    首先,判断可不可以只需要把原来的行列数相乘如果等于新的则可以,否则不行。

    其次,如果可以的话,只需要顺序读取原来的数组放到合适的位置即可。或者数序循环新的数组,每个位置放上合适的元素即可

    public int[][] matrixReshape(int[][] nums, int r, int c) {
            if(nums.length*nums[0].length!=r*c)
            {
                return nums;
            }
            else
            {
                int[][] result = new int[r][c];
                int orignRow = nums.length;
                int orignCloum = nums[0].length;
                int tempRow = 0;
                int tempCloum = 0;
                for(int i=0;i<r;i++)
                {
                    for(int j=0;j<c;j++)
                    {
                        if(tempCloum<orignCloum)
                        {
                            result[i][j] = nums[tempRow][tempCloum];
                            tempCloum++;
                        }
                        else
                        {
                            tempCloum = 0;
                            tempRow++;
                            result[i][j] = nums[tempRow][tempCloum];
                            tempCloum++;
                        }
                    }
                }
                return result;
            }
        }
    }
  • 相关阅读:
    某一字段分组取最大(小)值所在行的数据
    【JVM】01虚拟机内存模型
    POJ 1845 Sumdiv (求某个数的所有正因子的和)
    POJ 2992 Divisors (求因子个数)
    POJ 3696 The Luckiest number (欧拉函数,好题)
    POJ 1811 Prime Test (Pollard rho 大整数分解)
    POJ 2429 GCD & LCM Inverse (Pollard rho整数分解+dfs枚举)
    POJ 1305 Fermat vs. Pythagoras (毕达哥拉斯三元组)
    POJ 2142 The Balance (解不定方程,找最小值)
    POJ 1006 Biorhythms (中国剩余定理)
  • 原文地址:https://www.cnblogs.com/icysnow/p/8206574.html
Copyright © 2011-2022 走看看