package main
import "fmt"
// 执行N-1轮
// 第1轮找到最大值
// 第N-1轮找到次小值
func bubbleSort(nums []int) []int {
for i := 0; i < len(nums)-1; i++ {
for j := 0; j < len(nums)-1-i; j++ {
if nums[j] > nums[j+1] {
nums[j], nums[j+1] = nums[j+1], nums[j]
}
}
}
return nums
}
func main() {
var nums = []int{5, 2, 7, 4, 3, 6}
fmt.Println(nums)
fmt.Println(bubbleSort(nums))
}
package main
import "fmt"
func quickFind(list []string) bool {
var input string
fmt.Println("请输入一个名字:")
fmt.Scanln(&input)
for _, item := range list {
if input == item {
return true
}
}
return false
}
func main() {
var list = []string{"白眉鹰王", "金毛狮王", "紫衫龙王", "青翼福王"}
fmt.Println(quickFind(list))
}
package main
import "fmt"
// 二分查找,最多找len(list)/2轮
// 二分查找,要求数组有序
func binaryFind(list []int, num int) int {
var index = -1
var start = 0
var mid = len(list) / 2
var end = len(list) - 1
for i := 0; i <= len(list)/2; i++ {
mid = (start + end) / 2
if list[mid] == num {
return mid
}
if list[mid] > num {
end = mid - 1
continue
}
if list[mid] < num {
start = mid + 1
continue
}
}
return index
}
func main() {
var list = []int{1, 8, 10, 89, 1000, 1234}
fmt.Println(list)
fmt.Println("请输入一个要查找的数值:")
var num int
fmt.Scanln(&num)
index := binaryFind(list, num)
if index == -1 {
fmt.Println("没有这个数")
} else {
fmt.Println(index)
}
}
package main
import "fmt"
func main() {
// for i := 0; i < 4; i++ {
// for j := 0; j < 6; j++ {
// if i == 1 && j == 2 {
// fmt.Print(1)
// } else if i == 2 && j == 1 {
// fmt.Print(2)
// } else if i == 2 && j == 3 {
// fmt.Print(3)
// } else {
// fmt.Print(0)
// }
// }
// fmt.Println()
// }
var demArray [4][6]int
demArray[1][2] = 1
demArray[2][1] = 2
demArray[2][3] = 3
for i := 0; i < 4; i++ {
for j := 0; j < 6; j++ {
fmt.Print(demArray[i][j])
}
fmt.Println()
}
}
package main
import "fmt"
func main() {
var demArray = [3][5]int{
{50, 60, 70, 80, 90},
{51, 61, 71, 81, 91},
{52, 62, 72, 82, 92},
}
var sum int
for i := 0; i < 3; i++ {
var subSum int
for j := 0; j < 5; j++ {
sum += demArray[i][j]
subSum += demArray[i][j]
}
fmt.Println(subSum / 5)
}
fmt.Println(sum / (3 * 5))
}