726. Number of Atoms

原题链接

思路

思路1

设置两个栈，一个放括号，另一个放Map，放一个左括号，就放一个Map，括号出栈时，把另一个栈头的map弹出，值加入到新的栈顶map。

public String countOfAtoms(String formula) {
	Deque stack1 = new ArrayDeque<>();
    Deque> stack2 = new ArrayDeque<>();

    Map map = new HashMap<>();
    stack2.push(map);

    for (int i = 0; i < formula.length(); ) {
        char c = formula.charAt(i);

        if (c == '(') {
            stack1.push('(');
            stack2.push(new HashMap<>());
            i++;
            continue;
        }

        if (c == ')') {
            int num = 0;
            int j = i+1;
            for (; j < formula.length(); j++) {
                char c1 = formula.charAt(j);
                if (Character.isDigit(c1)) {
                    num = num * 10 + Character.getNumericValue(c1); // bug2 = 写为了 +=
                } else {
                    break;
                }
            }
            i = j;

            stack1.pop();
            Map poll = stack2.poll();
            Map peek = stack2.peek();
            for (String key: poll.keySet()) {
                if (peek.containsKey(key)) {
                    peek.put(key, peek.get(key) + poll.get(key) * num);
                } else {
                    peek.put(key, poll.get(key) * num); // bug1 没有乘以num
                }
            }

            continue;
        }

        if (Character.isUpperCase(c)) {
            String s = Character.toString(c);
            int num = 0;

            int j = i+1;
            for (; j < formula.length(); j++) {
                char c1 = formula.charAt(j);
                if (Character.isLowerCase(c1)) {
                    s += Character.toString(c1);
                } else {
                    break;
                }
            }

            for (; j < formula.length(); j++) {
                char c1 = formula.charAt(j);
                if (Character.isDigit(c1)) {
                    num = num * 10 + Character.getNumericValue(c1);
                } else {
                    break;
                }
            }

            num = num == 0 ? 1 : num;

            Map peek = stack2.peek();
            if (peek.containsKey(s)) {
                peek.put(s, peek.get(s) + num);
            } else {
                peek.put(s, num);
            }

            i = j;
            continue;
        }

        System.out.println("出错了！" + i);
    }

    Map pop = stack2.pop();

    List list = new ArrayList<>();
    list.addAll(pop.keySet());
    Collections.sort(list);

    StringBuffer sb = new StringBuffer();
    for (String k: list) {
        Integer v = pop.get(k);
        sb.append(k);
        if (v > 1) {
            sb.append(v);
        }
    }

    return sb.toString();
}

优化

看了评论区的答案，发现存在括号的栈没有必要存在，可以去掉。

总结

思路正确，过程有两个bug。
1、没有乘以num；
2、= 写成了 +=