题目描述

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given “25525511135”,

return [“255.255.11.135”, “255.255.111.35”]. (Order does not matter)

代码

    List<String> result = new ArrayList<String>();
    String[] stack = new String[4];

    public List<String> restoreIpAddresses(String s) {

        if (s == null || s.length() == 0) {
            return new ArrayList<String>();
        }
        dfs(s, 0, 0);
        return result;
    }

    /** * * @param s * @param p * :指针 * @param pstack * :stack的下标 */
    public void dfs(String s, int p, int pstack) {

        if (pstack == 4) {

            // 如果stack长度为4，且s的字符全部用上
            // 则stack[0...3]存了一个结果
            if (p >= s.length()) {
                String ip = String.join(".", stack);
                result.add(ip);
            }
            return;
        }

        // 获取1~3个字符
        for (int i = 1; i <= 3; i++) {

            // 如果超过字符串长度，返回
            if (p + i > s.length()) {
                return;
            }

            // 若选取的第一个字符是0，则停止选取
            if (i > 1 && s.charAt(p) == '0') {
                continue;
            }

            String number = s.substring(p, p + i);

            // 如果number<=255，递归
            if (Integer.parseInt(number) <= 255) {
                stack[pstack] = number;
                dfs(s, p + i, pstack + 1);
            }
        }
    }