93. Restore IP Addresses 93.恢复IP地址

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

A valid IP address consists of exactly four integers (each integer is between 0 and 255) separated by single points.

Example:

Input: "25525511135"
Output: ["255.255.11.135", "255.255.111.35"]


原来的思路：

    //这里不是从1开始的，是从start开始的，然后i最多等于start + 3.
    //backtrace中，start递归换成i + 1
    for (int sectionLength = 1; sectionLength < 4; sectionLength++) {
        if (isValid())
    }

应该有的思路：其实不用纠结什么字符串长度123，保证每一截都是有效的就行了。head这一截和剩下的部分。

控制remainingCount从4开始即可。

退出条件：s.length() == 0，不是s == ""

从1到最后一位，因为取head的时候需要包括在内
for (int i = 1; i <= s.length(); i++)

head是valid就回溯，不行就break

判断第一位是否等于0的方法：看看字符串长度和数字长度是否相等

s.length() == String.valueOf(val).length()

class Solution {
    //右边需要初始化
    List<String> results = new ArrayList<>();
    
    public List<String> restoreIpAddresses(String s) {
        //cc
        if (s.length() == 0) 
            return results;
        
        backtrace(s, "", 4);
        
        return results;
    }
    
    public void backtrace(String s, String currentString, int remainingCount) {
        //exit
       if (s.length() == 0 && remainingCount == 0) {
           results.add(currentString.substring(0, currentString.length() - 1));
           return ;
       }
        
        if (s.length() == 0 || remainingCount == 0) 
            return ;
        
        for (int i = 1; i <= s.length(); i++) {
            String head = s.substring(0, i);
            
            if (isValid(head)) {
                backtrace(s.substring(i), currentString + head + ".", 
                         remainingCount - 1);
            }else break;
        }
        
    }
    
    private boolean isValid(String s) {
    int val = Integer.valueOf(s);
    if (val > 255) return false;
    return s.length() == String.valueOf(val).length();
}
}