LeetCode: Restore IP Addresses 解题报告

Restore IP Addresses

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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)

SOLUTION 1:

如果有跟过主页君的讲解，就会知道，这又是一道相当经典的DFS模板题目。 我们照样套用http://www.ninechapter.com/的递归

模板，退出条件是：path.size == 4.

在模板中，我们加入一些判断条件来中断循环：例如说判断pre的字符串转化后的数字是不是在255以内。

另外，我们要排除055这种数字，所以加入这一行判断：

if (s.charAt(index) == '0' && i > index) {
　　break;
}

虽然是很简单的递归题，但主页君是真心用心写了的。而且这是相当经典的递归模板。同学们一定要记住这种模板解法哦！

public List<String> restoreIpAddresses(String s) {
        if (s == null) {
            return null;
        }
        
        ArrayList<String> ret = new ArrayList<String>();
        ArrayList<String> path = new ArrayList<String>();
        
        dfs(s, 0, path, ret);
        
        return ret;
    }
    
    public void dfs(String s, int index, ArrayList<String> path, ArrayList<String> ret) {
        if (path.size() == 4) {
            if (index == s.length()) {
                StringBuilder sb = new StringBuilder();
                for (String str: path) {
                    sb.append(str);
                    sb.append('.');
                }
                
                sb.deleteCharAt(sb.length() - 1);
                ret.add(sb.toString());
            }
            
            return;
        }
        
        int len = s.length();
        for (int i = index; i < index + 3 && i < len; i++) {
            if (s.charAt(index) == '0' && i > index) {
                break;
            }
            
            String pre = s.substring(index, i + 1);
            int num = Integer.parseInt(pre);
            if (num > 255) {
                continue;
            }
            
            path.add(pre);
            dfs(s, i + 1, path, ret);
            path.remove(path.size() - 1);
        }
    }

2015.1.1 redo:

容易出错的点：1. i的索引，注意不要越界。2. 记得把sb添加到ret中。

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> ret = new ArrayList<String>();
        // Bug 1: not length, but length().
        if (s == null || s.length() < 4 || s.length() > 12) {
            return ret;
        }
        
        dfs(s, new ArrayList<String>(), ret, 0);
        return ret;
    }
    
    public void dfs(String s, List<String> path, List<String> ret, int index) {
        // THE BASE CASE:
        int len = s.length();
        if (path.size() == 4) {
            // Create a solution.
            if (index == len) {
                StringBuilder sb = new StringBuilder();
                for (String str: path) {
                    sb.append(str);
                    sb.append(".");
                }
                sb.deleteCharAt(sb.length() - 1);
                
                // bug 3: forget this statement.
                ret.add(sb.toString());
            }
            
            return;
        }
        
        // 2/ 25 / 255
        // bug 2: i should < len.
        for (int i = index; i < index + 3 && i < len; i++) {
            String sub = s.substring(index, i + 1);
            if (s.charAt(index) == '0' && i != index) {
                // only allow 0, not 02, 022
                break;
            }
            
            if (!isValid(sub)) {
                continue;
            }
            
            path.add(sub);
            dfs(s, path, ret, i + 1);
            path.remove(path.size() - 1);
        }
    }
    
    public boolean isValid(String s) {
        int num = Integer.parseInt(s);
        return num >= 0 && num <= 255;
    }
}

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/RestoreIpAddresses.java