zoukankan      html  css  js  c++  java
  • Leetcode刷题记录[java]——566 Reshape the Matrix

    一、前言

    二、题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.

    三、解题思路

      用新的行数、列数reshape二维数组,数组元素的值不变。

      有一个技巧:遍历数组元素时,建立新数组与原数组的对应关系。

     1 class Solution {
     2     public int[][] matrixReshape(int[][] nums, int r, int c) {    
     3         int n = nums.length, m = nums[0].length;
     4         if (r*c != n*m) return nums;
     5         int[][] res = new int[r][c];
     6         for (int i=0;i<r*c;i++) 
     7             res[i/c][i%c] = nums[i/m][i%m];
     8         return res;       
     9     }
    10 }
  • 相关阅读:
    LCA + 二分(倍增)
    Educational Codeforces Round 5
    BNU 51276
    POJ 1511
    hdu2121
    最小树形图(朱刘算法)
    Educational Codeforces Round 1(D. Igor In the Museum) (BFS+离线访问)
    Educational Codeforces Round 1(C. Nearest vectors)
    POJ-2785 4 Values whose Sum is 0(折半枚举 sort + 二分)
    POJ 1661Help Jimmy(逆向DP Or 记忆化搜索 Or 最短路径)
  • 原文地址:https://www.cnblogs.com/Myoungs/p/7833050.html
Copyright © 2011-2022 走看看