golang顺时针打印矩阵的方法示例
2019/7/10 22:04:20
本文主要是介绍golang顺时针打印矩阵的方法示例,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
题目描述
输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字,例如,如果输入如下4 X 4矩阵: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.
package main import "fmt" func main() { //s := [][]int{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}} //s := [][]int{{1}, {2}, {3}, {4}} //s := [][]int{{1, 2, 3, 4}, {5, 6, 7, 8}} s := [][]int{{1, 2}, {5, 6}, {9, 10}, {13, 14}} printMatrix(s) } func printMatrix(s [][]int) { if s == nil { fmt.Println("切片为空,无法打印") } bex, bey := 0, 0 hang := len(s) - 1 lie := len(s[0]) - 1 if hang == 0 { for _, v := range s[0] { fmt.Println(v) } return } if lie == 0 { for _, v := range s { fmt.Println(v[0]) } return } for bex <= hang && bey <= lie { ax, ay := bex, bey for ay < lie { fmt.Println(s[ax][ay]) ay++ } for ax < hang { fmt.Println(s[ax][ay]) ax++ } for ay > bey { fmt.Println(s[ax][ay]) ay-- } for ax > bex { fmt.Println(s[ax][ay]) ax-- } bex++ bey++ hang-- lie-- } }
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