[用Golang刷LeetCode之 7] 566. Reshape the Matrix

miltonsun · · 630 次点击 · · 开始浏览    
这是一个创建于 的文章,其中的信息可能已经有所发展或是发生改变。

题目

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.

题目大意:

给定二维矩阵nums,将其转化为r行c列的新矩阵。若无法完成转化,返回原矩阵。

注意:

给定矩阵的高度和宽度范围[1, 100]
r和c都是正数

解题思路

  1. origin_r * origin_c 的矩阵 reshape为 rc的矩阵,需要满足:
    origin_r * origin_c=r
    c
  2. 元素位置对应的关系
    如果将矩阵横向展开为一维数组,元素个数为n=origin_r * origin_c
    在元素在一维数组中对应的位置i:
    原矩阵位置[i/origin_c,i%origin_c]
    新矩阵位置[i/c,i%c]

代码

reshapeMatrix.go

package _566_Reshape_Matrix

import "fmt"

func MatrixReshape(nums [][]int, r int, c int) [][]int {
    var chanInt chan int
    chanInt = make(chan int)

    var length int
    go func(len *int) {
        for _, v1 := range nums {
            for _, v := range v1 {
                (*len)++
                fmt.Printf("len1:%+v\n", length)
                chanInt <- v
            }
        }

        for {
            chanInt <- 0
        }
    }(&length)


    var ret [][]int
    for i := 0; i < r; i++ {
        var lineRet []int
        for j := 0; j < c; j++ {
            v := <- chanInt

            lineRet = append(lineRet, v)
        }
        ret = append(ret, lineRet)
    }

    fmt.Printf("len2:%+v\n", length)
    fmt.Printf("ret:%+v\n", ret)
    if r * c != length {
        return nums
    }

    return ret
}

测试

reshapeMatrix_test.go

package _566_Reshape_Matrix

import "testing"

func TestMatrixReshape(t *testing.T) {
    input := [][]int{
        {1, 2},
        {3, 4},
    }
    ret := MatrixReshape(input, 1, 4)

    t.Logf("%+v", ret)
}


有疑问加站长微信联系(非本文作者)

本文来自:简书

感谢作者:miltonsun

查看原文:[用Golang刷LeetCode之 7] 566. Reshape the Matrix

入群交流(和以上内容无关):加入Go大咖交流群,或添加微信:liuxiaoyan-s 备注:入群;或加QQ群:692541889

630 次点击  
加入收藏 微博
暂无回复
添加一条新回复 (您需要 登录 后才能回复 没有账号 ?)
  • 请尽量让自己的回复能够对别人有帮助
  • 支持 Markdown 格式, **粗体**、~~删除线~~、`单行代码`
  • 支持 @ 本站用户;支持表情(输入 : 提示),见 Emoji cheat sheet
  • 图片支持拖拽、截图粘贴等方式上传