Number of Atoms

最新推荐文章于 2019-05-22 04:27:45 发布

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本文介绍了一种解析化学分子式并计算各元素数量的算法。通过使用栈和递归方法，该算法能够正确处理嵌套括号，实现对复杂分子式的精确解析。

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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.

题目理解：

给定分子式，计算分子式中每一种元素的数量

解题思路：

这个题目主要的难点在于如何处理括号，因为同一种元素可能出现在括号里面和外面，在计算括号中元素数量时，需要对这两部分的元素分别处理。

我采用的方法是将元素的种类和数量作为一个整体存入栈中，将所有的左括号“（”直接入栈，如果遇到有括号“）”，那么将栈中左括号之前的所有元素取出来，改写其数量，再重新加入栈中，这样就可以解决不同位置的同种元素的数量处理问题，同时也可以处理括号的嵌套

class Solution {
    public String countOfAtoms(String formula) {
        char[] chs = formula.toCharArray();
        int it = 0, len = formula.length();
        Stack<String> st = new Stack<>();
        while(it < len){
            char ch = chs[it];
            if(ch >= 'A' && ch <= 'Z'){
                String atom = "" + ch;
                it++;
                while(it < len && chs[it] >= 'a' && chs[it] <= 'z'){
                    atom += chs[it];
                    it++;
                }
                st.push(atom + ",1");
            }
            else if(ch >= '0' && ch <= '9'){
                String num = "" + ch;
                it++;
                while(it < len && chs[it] >= '0' && chs[it] <= '9'){
                    num += chs[it];
                    it++;
                }
                int count = Integer.valueOf(num);
                if(st.peek().equals(")")){
                    List<String> list = new ArrayList<>();
                    st.pop();
                    while(!st.peek().equals("("))
                        list.add(st.pop());
                    st.pop();
                    for(String atom : list) {
                        int pos = atom.indexOf(',');
                        int base = Integer.valueOf(atom.substring(pos + 1));
                        int cur = base * count;
                        st.push(atom.substring(0, pos) + "," + cur);
                    }
                }
                else{
                    String atom = st.pop();
                    int pos = atom.indexOf(',');
                    int base = Integer.valueOf(atom.substring(pos + 1));
                    int cur = base * count;
                    st.push(atom.substring(0, pos) + "," + cur);
                }
            }
            else if(ch == '(' || ch == ')'){
                st.push(String.valueOf(ch));
                it++;
            }
        }
        TreeMap<String, Integer> map = new TreeMap<>();
        for(String str : st){
            int pos = str.indexOf(',');
            String atom = str.substring(0, pos);
            int num = Integer.valueOf(str.substring(pos + 1));
            map.put(atom, map.getOrDefault(atom, 0) + num);
        }
        StringBuilder sb = new StringBuilder("");
        for(String atom : map.keySet()){
            sb.append(atom);
            int num = map.get(atom);
            if(num > 1)
                sb.append(map.get(atom));
        }
        return sb.toString();
    }
}