726. Number of Atoms #
Problem #
Given a chemical formula (given as a string), return the count of each atom.
An atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name.
1 or more digits representing the count of that element may follow if the count is greater than 1. If the count is 1, no digits will follow. For example, H2O and H2O2 are possible, but H1O2 is impossible.
Two formulas concatenated together produce another formula. For example, H2O2He3Mg4 is also a formula.
A formula placed in parentheses, and a count (optionally added) is also a formula. For example, (H2O2) and (H2O2)3 are formulas.
Given a formula, output the count of all elements as a string in the following form: the first name (in sorted order), followed by its count (if that count is more than 1), followed by the second name (in sorted order), followed by its count (if that count is more than 1), and so on.
Example 1:
Input:
formula = "H2O"
Output: "H2O"
Explanation:
The count of elements are {'H': 2, 'O': 1}.
Example 2:
Input:
formula = "Mg(OH)2"
Output: "H2MgO2"
Explanation:
The count of elements are {'H': 2, 'Mg': 1, 'O': 2}.
Example 3:
Input:
formula = "K4(ON(SO3)2)2"
Output: "K4N2O14S4"
Explanation:
The count of elements are {'K': 4, 'N': 2, 'O': 14, 'S': 4}.
Note:
- All atom names consist of lowercase letters, except for the first character which is uppercase.
- The length of
formulawill be in the range[1, 1000]. formulawill only consist of letters, digits, and round parentheses, and is a valid formula as defined in the problem.
Problem Summary #
Given a chemical formula, output the count of all atoms. The format is: the name of the first atom (in lexicographical order), followed by its count (if the count is greater than 1), then the name of the second atom (in lexicographical order), followed by its count (if the count is greater than 1), and so on.
An atom always starts with an uppercase letter, followed by 0 or any number of lowercase letters, representing the atom’s name. If the count is greater than 1, digits follow the atom to represent its count. If the count equals 1, no digits follow. For example, H2O and H2O2 are valid, but H1O2 is not valid. Two chemical formulas concatenated together form a new chemical formula. For example, H2O2He3Mg4 is also a chemical formula. A chemical formula inside parentheses followed by a number (optionally added) is also a chemical formula. For example, (H2O2) and (H2O2)3 are chemical formulas.
Solution Ideas #
- Use a stack to process each chemical element, use a map to record the count of each chemical element, and finally sort and output the result
- Note that some chemical elements are not single letters, such as magnesium, which is Mg, so letter case needs to be considered.
Code #
package leetcode
import (
"sort"
"strconv"
"strings"
)
type atom struct {
name string
cnt int
}
type atoms []atom
func (this atoms) Len() int { return len(this) }
func (this atoms) Less(i, j int) bool { return strings.Compare(this[i].name, this[j].name) < 0 }
func (this atoms) Swap(i, j int) { this[i], this[j] = this[j], this[i] }
func (this atoms) String() string {
s := ""
for _, a := range this {
s += a.name
if a.cnt > 1 {
s += strconv.Itoa(a.cnt)
}
}
return s
}
func countOfAtoms(s string) string {
n := len(s)
if n == 0 {
return ""
}
stack := make([]string, 0)
for i := 0; i < n; i++ {
c := s[i]
if c == '(' || c == ')' {
stack = append(stack, string(c))
} else if isUpperLetter(c) {
j := i + 1
for ; j < n; j++ {
if !isLowerLetter(s[j]) {
break
}
}
stack = append(stack, s[i:j])
i = j - 1
} else if isDigital(c) {
j := i + 1
for ; j < n; j++ {
if !isDigital(s[j]) {
break
}
}
stack = append(stack, s[i:j])
i = j - 1
}
}
cnt, deep := make([]map[string]int, 100), 0
for i := 0; i < 100; i++ {
cnt[i] = make(map[string]int)
}
for i := 0; i < len(stack); i++ {
t := stack[i]
if isUpperLetter(t[0]) {
num := 1
if i+1 < len(stack) && isDigital(stack[i+1][0]) {
num, _ = strconv.Atoi(stack[i+1])
i++
}
cnt[deep][t] += num
} else if t == "(" {
deep++
} else if t == ")" {
num := 1
if i+1 < len(stack) && isDigital(stack[i+1][0]) {
num, _ = strconv.Atoi(stack[i+1])
i++
}
for k, v := range cnt[deep] {
cnt[deep-1][k] += v * num
}
cnt[deep] = make(map[string]int)
deep--
}
}
as := atoms{}
for k, v := range cnt[0] {
as = append(as, atom{name: k, cnt: v})
}
sort.Sort(as)
return as.String()
}
func isDigital(v byte) bool {
if v >= '0' && v <= '9' {
return true
}
return false
}
func isUpperLetter(v byte) bool {
if v >= 'A' && v <= 'Z' {
return true
}
return false
}
func isLowerLetter(v byte) bool {
if v >= 'a' && v <= 'z' {
return true
}
return false
}
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