前言

关于字符串更多的是梳理清楚逻辑，将自己的思路完整的表现出来，刷起来吧！

有效的括号

看到字符匹配，第一时间就想到通过栈这个特殊的数据结构进行匹配。

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack();
        Map<Character, Character> map = new HashMap<Character, Character>();
        map.put(')','(');
        map.put('}','{');
        map.put(']','[');
        String in = "({[";
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(!map.containsKey(c)){
                stack.push(c);
            }else{
                if(stack.isEmpty()||stack.peek()!=map.get(c)){
                    return false;
                }
                stack.pop();
            }   
        }
        return stack.isEmpty();
    }
}

括号生成

这到题是上一道「有效括号」的进阶，可以先对所有可能做一个全排列，然后依次判断是否符合条件。

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> combinations = new ArrayList<String>();
        generateAll(new char[2 * n], 0, combinations);
        return combinations;
    }

    public void generateAll(char[] current, int pos, List<String> result) {
        if (pos == current.length) {
            if (valid(current)) {
                result.add(new String(current));
            }
        } else {
            current[pos] = '(';
            generateAll(current, pos + 1, result);
            current[pos] = ')';
            generateAll(current, pos + 1, result);
        }
    }

    public boolean valid(char[] current) {
        int balance = 0;
        for (char c: current) {
            if (c == '(') {
                ++balance;
            } else {
                --balance;
            }
            if (balance < 0) {
                return false;
            }
        }
        return balance == 0;
    }
}

二进制求和

class Solution {
    public String addBinary(String a, String b) {
        StringBuffer ans = new StringBuffer();

        int n = Math.max(a.length(), b.length()), carry = 0;
        for (int i = 0; i < n; ++i) {
            carry += i < a.length() ? (a.charAt(a.length() - 1 - i) - '0') : 0;
            carry += i < b.length() ? (b.charAt(b.length() - 1 - i) - '0') : 0;
            ans.append((char) (carry % 2 + '0'));
            carry /= 2;
        }

        if (carry > 0) {
            ans.append('1');
        }
        ans.reverse();

        return ans.toString();
    }
}