用golang写这道题的做法完全不同,用到了gorutine, channel,写着挺有意思的
题目描述:
LeetCode 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 crepresenting 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:
The height and width of the given matrix is in range [1, 100].
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:
详见代码
代码
reshapeMatrix.go
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
}
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