我正在尝试查找在字母的圆形排列中均等分开的单词/字符串。例如:
因此,我想编写函数IsSeparated(B),如果B是“ isSeparated”,则返回true
以下是我的代码/解决方案:
maxpair -1 == count
因为总会有1个字母没有对[{ab} {bx} {xy} {yz} {za}] - [{0} {21} {0} {0} {0}]]//there are 5 pairs = maxPair -1({-xy}
因此,由于它是圆形排列的,所以中间的一个将始终是奇数,即21,与其余的对不相等地分开
这是棘手的部分,我似乎无法获得所需的输出。找到字母顺序的每个字母的长度/分隔并检查它们是否均匀分隔的正确方法是什么?
package main
import (
"fmt"
"strings"
)
//Q3
func separationCount(x, y string) int {
alphabets := [26]string{"a","b","c","d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u","v", "w", "x", "y", "z"}
separation := 0
for i:=0; i < len(alphabets); i++{
if x == alphabets[i]{
for j:= i+1; j <len(alphabets); j++{
if y == alphabets[i+1]{
fmt.Println(separation)
return separation
}else{
i++
separation++
}
}
}else{
//do nothing
}
}
//fmt.Println(separation)
return 0
}
func isSeparated(B [] string) bool {
var N int = len(B) - 1
var maxPair int
var item1 string
var item2 string
var separation int = 0
count := 0
var intialSeparation int
//calling the methods
fmt.Println("Original array:",B)
B = removeDuplicates(B)
B = sortedList(B)
item1 = B[0]
item2 = B[1]
intialSeparation = separationCount(item1,item2)
for i := 0; i< N; i++{
item1 = B[i]
item2 = B[i + 1]
separation = separationCount(item1,item2)
maxPair++
if intialSeparation == separation{
count++
}
if maxPair == count{
return true
}else{
return false
}
}
return false
}
//to sort the alphabets
func sortedList(B []string) [] string {
N := len(B)
//max := 0
element1 := 0
element2 := 1
for element2 < N {
var item1 string = B[element1]
var item2 string = B[element2]
//using function call
if greater(item1, item2){
B[element1] = item2
B[element2] = item1
}
element1++
element2++
}
fmt.Println("Alphabetically sorted:", B )
return B
}
//for sorting
func greater(a, b string) bool {
if strings.ToLower(a) > strings.ToLower(b) {
return true
} else {
return false
}
}
//removing duplicates
func removeDuplicates(B []string) []string {
encountered := map[string]bool{}
// Create a map of all unique elements.
for v:= range B {
encountered[B[v]] = true
}
// Place all keys from the map into a slice.
result := []string{}
for key, _ := range encountered {
result = append(result, key)
}
fmt.Println("Duplicates removed:", result )
return result
}
func main(){
//q3
B := []string{"y", "a", "a", "a", "c", "e", "g", "w", "w", "w"}
fmt.Println(isSeparated(B))
}
我不太了解您尝试确定分离的部分。在Go中,就像在C中一样,您可以对字符进行算术运算。例如,您将获得每个小写字母的从0开始的索引:
pos := char - 'a';
你可以转向"abxyz"
到
{0, 1, 23, 24, 25}.
如果您将相邻字母之间的差值考虑在内,则会得到
{-25, 1, 22, 1, 1}
(-25 ist表示最后一个值与第一个值之间的差。)您有两个间隙:一个间隙,循环在b和w之间开始,另一个间隙,在字母换行之间。第二个差距是差异为负,始终在最后一项和第一项之间。您可以在差值上加上26来进行调整,也可以使用模数算法,在其中使用余数%
来考虑换行:
diff := ((p - q + 26) % 26;
的%
力的结果到范围从0到25,如果第一操作数是正的。+ 26表示它是肯定的。(下面的程序使用25,因为您对分离的定义不是位置的差异,而是两者之间的差异。)
现在你有差异
{1, 1, 22, 1, 1}
如果您最多只有两个不同的值,并且其中一个最多出现一次,则满足您的条件。(这是一个条件,我发现要测试它出奇地复杂,请参阅下文,但这部分是因为Go的地图有点笨拙。)
无论如何,这是代码:
package main
import "fmt"
func list(str string) int {
present := [26]bool{}
pos := []int{}
count := map[int]int{}
// determine which letters exist
for _, c := range str {
if 'a' <= c && c <= 'z' {
present[c-'a'] = true
}
}
// concatenate all used letters (count sort, kinda)
for i := 0; i < 26; i++ {
if present[i] {
pos = append(pos, i)
}
}
// find differences
q := pos[len(pos)-1]
for _, p := range pos {
diff := (p - q + 25) % 26
count[diff]++
q = p
}
// check whether input is a "rambai"
if len(count) > 2 {
return -1
}
which := []int{}
occur := []int{}
for k, v := range count {
which = append(which, k)
occur = append(occur, v)
}
if len(which) < 2 {
return which[0]
}
if occur[0] != 1 && occur[1] != 1 {
return -1
}
if occur[0] == 1 {
return which[1]
}
return which[0]
}
func testme(str string) {
fmt.Printf("\"%s\": %d\n", str, list(str))
}
func main() {
testme("zzzzyyyybbbzzzaaaaaxxx")
testme("yacegw")
testme("keebeebheeh")
testme("aco")
testme("naan")
testme("mississippi")
testme("rosemary")
}
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