LWC 58：726. Number of Atoms
LWC 58：726. 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 formula will be in the range [1, 1000].
Formula will only consist of letters, digits, and round parentheses, and is a valid formula as defined in the problem.
思路： 递归真的是个好东西，思路比较暴力，如果没有遇到括号，则按照传统的检测字符和数字进行更新即可。如果检测到括号，则先得到括号内的元素和个数的映射，接着乘上大括号外的第一个num，再将第三部分的答案组合进来即可。
比如：
H2O2(SO2)23(OH)23
上述字符串可以组成三部分
lf = "H2O2"
mid = "(SO2)23"
rt = "(OH)23"
只有mid需要特殊处理，把{s = 1 * 32, O = 2 * 23 = 46}
而一旦解决mid，则rt跟着被解决，lf在之前就被解决了，具体看代码，不解释了。代码如下：
public String countOfAtoms(String formula) {
Map<String, Integer> map = go(formula.toCharArray(), 0, formula.length() - 1);
StringBuilder sb = new StringBuilder();
List<String> set = new ArrayList<>(map.keySet());
Collections.sort(set);
for (String key : set) {
sb.append(key);
if (map.get(key) > 1) {
sb.append(map.get(key));
}
}
return sb.toString();
}
public Map<String, Integer> go(char[] cs, int s, int e){
if (s > e) return new HashMap<>();
Map<String, Integer> ans = new HashMap<>();
int idx = new String(cs).substring(s, e + 1).indexOf("(");
if (idx != -1) idx += s;
if (idx == -1) {
StringBuilder sb = new StringBuilder();
for (int i = s; i <= e;) {
if (Character.isUpperCase(cs[i])) {
sb.append(cs[i]);
i ++;
if (i <= e && Character.isUpperCase(cs[i])) {
String ele = sb.toString();
if (!ans.containsKey(ele)) {
ans.put(ele, 1);
}
else {
ans.put(ele, ans.get(ele) + 1);
}
sb = new StringBuilder();
}
else if (i <= e && Character.isLowerCase(cs[i])) {
while (i <= e && Character.isLowerCase(cs[i])) {
sb.append(cs[i]);
i ++;
}
if (i <= e && !Character.isDigit(cs[i])){
String ele = sb.toString();
if (!ans.containsKey(ele)) {
ans.put(ele, 1);
}
else {
ans.put(ele, ans.get(ele) + 1);
}
sb = new StringBuilder();
}
}
if (i <= e && Character.isDigit(cs[i])) {
int num = 0;
while (i <= e && Character.isDigit(cs[i])) {
num = num * 10 + cs[i] - '0';
i ++;
}
String ele = sb.toString();
if (!ans.containsKey(ele)) {
ans.put(ele, num);
}
else {
ans.put(ele, ans.get(ele) + num);
}
sb = new StringBuilder();
}
}
}
String ele = sb.toString();
if (!ele.isEmpty()) {
if (!ans.containsKey(ele)) {
ans.put(ele, 1);
}
else {
ans.put(ele, ans.get(ele) + 1);
}
}
return ans;
}
else {
Map<String, Integer> lf = go(cs, s, idx - 1);
StringBuilder sb = new StringBuilder();
int p = 1;
Map<String, Integer> include;
int num = 0;
int j = -1;
for (int i = idx + 1; i <= e; ++i) {
sb.append(cs[i]);
if (cs[i] == '(') {
p ++;
}
else if (cs[i] == ')') {
p --;
if (p == 0) {
j = i + 1;
while (j <= e && Character.isDigit(cs[j])) {
num = num * 10 + cs[j] - '0';
j ++;
}
break;
}
}
}
String ele = sb.toString().substring(0, sb.length() - 1);
include = go(ele.toCharArray(), 0, ele.length() - 1);
for (String key : include.keySet()) {
include.put(key, include.get(key) * num);
}
Map<String, Integer> rt = go(cs, j, e);
add(ans, lf);
add(ans, include);
add(ans, rt);
return ans;
}
}
public void add(Map<String, Integer> ans, Map<String, Integer> tmp) {
for (String key : tmp.keySet()) {
if (!ans.containsKey(key)) {
ans.put(key, tmp.get(key));
}
else {
ans.put(key, ans.get(key) + tmp.get(key));
}
}
}注意一些细节： 首先，对于字符串”H2O2BeBe49”这样的formula，第一次一定会遇到大写字符，接着把大写字母后的小写字母全部给append上，这样只会遇到两种情况：下一个字符为数字，或者下一个字符为大写字符。
如过遇到字符，则计算num，如果遇到大写字符，则formula省略了1，需要加上。
代码更新如下：
public String countOfAtoms(String formula) {
Map<String, Integer> map = go(formula.toCharArray(), 0, formula.length() - 1);
StringBuilder sb = new StringBuilder();
List<String> set = new ArrayList<>(map.keySet());
Collections.sort(set);
for (String key : set) {
sb.append(key);
if (map.get(key) > 1) {
sb.append(map.get(key));
}
}
return sb.toString();
}
public Map<String, Integer> go(char[] cs, int s, int e){
if (s > e) return new HashMap<>();
Map<String, Integer> ans = new HashMap<>();
int idx = new String(cs).substring(s, e + 1).indexOf("(");
if (idx != -1) idx += s;
if (idx == -1) {
StringBuilder sb = new StringBuilder();
for (int i = s; i <= e;) {
if (Character.isUpperCase(cs[i])) {
sb.append(cs[i]);
i ++;
while (i <= e && Character.isLowerCase(cs[i])){
sb.append(cs[i]);
i ++;
}
if (i <= e && !Character.isDigit(cs[i])){
String ele = sb.toString();
if (!ans.containsKey(ele)) {
ans.put(ele, 1);
}
else {
ans.put(ele, ans.get(ele) + 1);
}
sb = new StringBuilder();
}
else if (i <= e && Character.isDigit(cs[i])) {
int num = 0;
while (i <= e && Character.isDigit(cs[i])) {
num = num * 10 + cs[i] - '0';
i ++;
}
String ele = sb.toString();
if (!ans.containsKey(ele)) {
ans.put(ele, num);
}
else {
ans.put(ele, ans.get(ele) + num);
}
sb = new StringBuilder();
}
}
}
String ele = sb.toString();
if (!ele.isEmpty()) {
if (!ans.containsKey(ele)) {
ans.put(ele, 1);
}
else {
ans.put(ele, ans.get(ele) + 1);
}
}
return ans;
}
else {
Map<String, Integer> lf = go(cs, s, idx - 1);
StringBuilder sb = new StringBuilder();
int p = 1;
Map<String, Integer> include;
int num = 0;
int j = -1;
for (int i = idx + 1; i <= e; ++i) {
sb.append(cs[i]);
if (cs[i] == '(') {
p ++;
}
else if (cs[i] == ')') {
p --;
if (p == 0) {
j = i + 1;
while (j <= e && Character.isDigit(cs[j])) {
num = num * 10 + cs[j] - '0';
j ++;
}
break;
}
}
}
String ele = sb.toString().substring(0, sb.length() - 1);
include = go(ele.toCharArray(), 0, ele.length() - 1);
for (String key : include.keySet()) {
include.put(key, include.get(key) * num);
}
Map<String, Integer> rt = go(cs, j, e);
add(ans, lf);
add(ans, include);
add(ans, rt);
return ans;
}
}
public void add(Map<String, Integer> ans, Map<String, Integer> tmp) {
for (String key : tmp.keySet()) {
if (!ans.containsKey(key)) {
ans.put(key, tmp.get(key));
}
else {
ans.put(key, ans.get(key) + tmp.get(key));
}
}
} 腾讯云开发者
