<p>A <strong>valid IP address</strong> consists of exactly four integers separated by single dots. Each integer is between <code>0</code> and <code>255</code> (<strong>inclusive</strong>) and cannot have leading zeros.</p>

<ul>
	<li>For example, <code>&quot;0.1.2.201&quot;</code> and <code>&quot;192.168.1.1&quot;</code> are <strong>valid</strong> IP addresses, but <code>&quot;0.011.255.245&quot;</code>, <code>&quot;192.168.1.312&quot;</code> and <code>&quot;192.168@1.1&quot;</code> are <strong>invalid</strong> IP addresses.</li>
</ul>

<p>Given a string <code>s</code> containing only digits, return <em>all possible valid IP addresses that can be formed by inserting dots into </em><code>s</code>. You are <strong>not</strong> allowed to reorder or remove any digits in <code>s</code>. You may return the valid IP addresses in <strong>any</strong> order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;25525511135&quot;
<strong>Output:</strong> [&quot;255.255.11.135&quot;,&quot;255.255.111.35&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000&quot;
<strong>Output:</strong> [&quot;0.0.0.0&quot;]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;101023&quot;
<strong>Output:</strong> [&quot;1.0.10.23&quot;,&quot;1.0.102.3&quot;,&quot;10.1.0.23&quot;,&quot;10.10.2.3&quot;,&quot;101.0.2.3&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 20</code></li>
	<li><code>s</code> consists of digits only.</li>
</ul>

class Solution {
public:
//有效IP不能有前导0，而且有数字范围，而且不能有特殊字符
//忽略的一点：需要有四段
//分割方式
    vector<string> res;

    bool isvaild(string t){
        if(t[0]=='0'&&t.size()>1) return false;
        for(char c:t){
            if(!isdigit(c)) return false;
        }
        int num=stoi(t);
        if(num>=0&&num<=255) return true;
        return false;
    }

    void traverse(string s,string& cnt,int start,int part){
        if(start==s.size()&&part==4){
            res.push_back(cnt);
            return;
        }
        for(int i=start;i<s.size()&&((i-start)<=2);++i){
            string t=s.substr(start,i-start+1);//截取长度
            if(!isvaild(t)) continue;
            //最后不加.
            if(part<3) t=t+".";
            string cnt_1=cnt;
            cnt+=t;
            traverse(s,cnt,i+1,part+1);
            cnt=cnt_1;
        }
    }

    vector<string> restoreIpAddresses(string s) {
        string t;
        traverse(s,t,0,0);
        return res;
    }
};
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