[Leetcode 74] 92 Restore IP Addresses

Problem:

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)

Analysis:

Originally, I thought DFS may be a good solution. But since the brute-force enumeration is not that complicated, So I tried brute force method. The key observation is that:

1. to restore the IP address, we need to divide the original string into four sub-parts, thus need to choose 3 positions in the string to divide it.

2. for each substring, we need to judge whether it's valid or not. The rule is:

　　a. the value of the substring should in the range of [0, 255]

　　b. if it's not a 0, then no leading 0 is allowed. thus "010", "01", "00", "000" are invalid.

3. to get the final IP address, we need only concatinate as follows: s1 + "." + s2 + "." + s3 + "." + s4

For the dividing part, we can make use of some knowledge of IP address to reduce the work:

1. length of IP address should between 4 and 12 inclusively. Any given string's length less than 4 or greater than 12 can not be used as an IP address.

2. each sub-part's length should between 1 and 4 inclusively. Thus each for loop only need to check 3 division positions.

Code:

class Solution {
public:
vector<string> restoreIpAddresses(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<string> res;

if (s.length() < 4 || s.length() > 12)
return res;

int len = s.length();

for (int i=1; i<4 && i<len; i++)
for (int j=i+1; j<i+4 && j<len; j++)
for (int k=j+1; k<j+4 && k<len; k++) {
if (isValid(s, 0, i) && isValid(s, i, j) &&
isValid(s, j, k) && isValid(s, k, len)) {
string tmp = s.substr(0, i) + "." +
s.substr(i, j-i) + "." +
s.substr(j, k-j) + "." +
s.substr(k, len-k);

res.push_back(tmp);
}

}

return res;
}

private:
//  check sub-string s[s, e)'s validity
bool isValid(string &s, int a, int e) {
//only 1 bit, true anyway
if (a+1 == e)
return true;

int val = 0;
bool leading = false;
for (int i=a; i<e; i++) {
if (leading) {
val = val * 10 + s[i] - '0';
} else {
if (s[i] == '0')
return false;
else {
val = val * 10 + s[i]-'0';
leading = true;
}
}
}

if (val<=255 && val>=0)
return true;
else
return false;
}
};

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