Number of Atoms——LeetCode进阶路

原题链接https://leetcode.com/problems/number-of-atoms/

题目描述

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.

思路分析

把给定化学式的原子数目分别统计出来，并返回原子和对应数目组成的字符串。
类似之前做的计算器，都是在于括号的嵌套，可以通过递归或者栈，其实两者思想上差不多啦

下面贴的题解是小陌看了花花酱大神之后的优化版，好多新特性自己还没有去使用的意识，还是用之前本方法，差距呐ヽ(。>д<)ｐ
here are花花酱大神http://zxi.mytechroad.com/blog/string/leetcode-726-number-of-atoms/

AC解

class Solution {
    private int i;
    public String countOfAtoms(String formula) {
        i = 0;
        StringBuilder res = new StringBuilder();
        Map<String,Integer> map = parse(formula);
        
        for(String s:map.keySet())
        {
            res.append(s);
            int number = map.get(s);
            if(number > 1)
            {
                res.append("" + number);
            }
        }
        
        return res.toString();        
    }
    
    public Map<String,Integer> parse(String f)
    {
        Map<String,Integer> tm = new TreeMap();
        while( i<f.length() && f.charAt(i) != ')')
        {
            if(f.charAt(i) == '(')
            {
                i ++;
                for(Map.Entry<String,Integer> tra : parse(f).entrySet())
                {
                    tm.put(tra.getKey(),tm.getOrDefault(tra.getKey(),0) + tra.getValue());
                }
            }
            else
            {
                int begin  = i ++;
                while(i<f.length() && Character.isLowerCase(f.charAt(i)))
                {
                    i ++;
                }
                
                String atom = f.substring(begin,i);
                begin = i;
                 while(i<f.length() && Character.isDigit(f.charAt(i)))
                {
                    i ++;
                }
                
                int count = begin < i ? Integer.parseInt(f.substring(begin,i)) : 1;
                tm.put(atom,tm.getOrDefault(atom,0) + count);
            }
        }
        
        int begin = ++ i;
        while(i<f.length() && Character.isDigit(f.charAt(i)))
        {
            i ++;
        }
        if(begin < i)
        {
            int count = Integer.parseInt(f.substring(begin,i));
            for(String s:tm.keySet())
            {
                tm.put(s,tm.get(s)*count);
            }
        }       
        
        return tm;
    }
}