Posts Tagged 【dfs】Restore IP Addresses

Restore IP Addresses

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Submissions: 145054My Submissions

Question
Solution

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)

/*
将字符串转化为4个数字，每个数字不能超过256，同时注意010
思路就是dfs：
每次三个选择：选择1个字符，2个字符还是3个字符
同时检查合法性
*/
public class Solution {
public List<String> restoreIpAddresses(String s) {
List<String> lists = new ArrayList<String>();
int[] startList = new int[5];
startList[0] = 0;
//第0次三种选择count=0
dfs(s,lists,0,1,0,startList);
dfs(s,lists,0,2,0,startList);
dfs(s,lists,0,3,0,startList);
return lists;

}
private void dfs(String s,List<String> lists,int start,int end,int count,int[] startList) {
if(count > 3) {
if(start != s.length()) return;
StringBuilder sb = new StringBuilder();
sb.append(s.substring(startList[0], startList[1]));
sb.append('.');
sb.append(s.substring(startList[1], startList[2]));
sb.append('.');
sb.append(s.substring(startList[2], startList[3]));
sb.append('.');
sb.append(s.substring(startList[3]));
lists.add(sb.toString());
return;
}
//count+1;
startList[++count] = end;
if(end <= s.length() && isVaild(s, start, end)) {
if(count < 4) {
//前4次选择count=1,2,3，包括0
dfs(s,lists,end,end+1,count,startList);
dfs(s,lists,end,end+2,count,startList);
dfs(s,lists,end,end+3,count,startList);
} else {
//4次都已选择，接下来dfs就是判断并打印。
//在判断是否打印中，由于判断的缘故，这里就一次
dfs(s,lists,end,end+1,count,startList);
}
} else {
return;
}
}
//判断是否合法
private boolean isVaild(String s,int start,int end) {
String sb = s.substring(start,end);
if(sb.isEmpty()) return false;
if(sb.matches("[0][0-9]+")) return false;
if( Integer.parseInt(sb)< 256)  return true;
return false;
}
}

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