面试题29：回文字符串

回文的题经常出现在算法题中，下面总结了leetcode出现的6个关于回文的题目。

1. 如何判断字符串是回文字符串？（easy）
2. 如何判断一个整数是回文的？（easy）
3. 如何用最少的切法将一个字符串切成几个回文串组成？（dp）
4. 寻找一个字符串中最长的回文串？（马拉车算法）
5. 如何通过在字符串的头部添加最少的字符，使得添加后的字符串成为回文字符串？（kmp中最长前缀后缀数组）
6. 将字符串划分为回文串，返回所有可能的结果？（dfs）

1.Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama"is a palindrome.
"race a car"is not a palindrome.

class Solution {
public:
    bool isAlphanumberic(char c){
        if((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') || (c >= '0' && c<='9')) return true;
        else return false;
    }
    bool isPalindrome(string s) {
        int n = s.length();
        int left = 0;
        int right = n - 1;
        while(left < right){
            if(!isAlphanumberic(s[left])){
                left++;
                continue;
            }

            if(!isAlphanumberic(s[right])){
                right--;
                continue;
            }
            if(s[left] == s[right] || abs(s[left] - s[right]) == 32){
                left++;
                right--;
            }else{
                return false;
            }
        }
        return true;
    }
};

2.

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

class Solution {
public:
    bool isPalindrome(int x) {
        if(x < 0) return false;
        int len = 1;
        while(x / len >= 10){
            len *= 10;
        }

        while(x > 0){
            int left = x / len;
            int right = x % 10;
            if(left != right){
                return false;
            }else{
                x = x % len / 10;
                len /= 100;
            }
        }
        return true;
    }
};

3.

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s ="aab",
Return1since the palindrome partitioning["aa","b"]could be produced using 1 cut.

class Solution {
public:
    bool isPalindrome(string s) {
        int n = s.length();
        if (n < 2)
            return true;
        int left = 0;
        int right = n - 1;
        while (left < right) {
            if (s[left] == s[right]) {
                left++;
                right--;
            } else {
                return false;
            }
        }
        return true;
    }
    int minCut(string s) {
        int n = s.length();

        vector<int> dp(n,n-1);
        for (int i = 1; i <= n; i++) {
            if (isPalindrome(s.substr(0, i)))
                dp[i-1] = 0;
            else {
                for (int j = 1; j < i; j++) {
                    if (isPalindrome(s.substr(j, i - j))) {
                        dp[i-1] = min(dp[i-1],dp[j-1] + 1);
                    }
                }
            }

        }
        return dp[n-1];
    }
};

4.Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

class Solution {
public:
    string longestPalindrome(string s) {
        string t = "$#";
        for (size_t i = 0; i < s.length(); i++) {
            t += s[i];
            t += "#";
        }

        int mx = 0;
        int id = 0;
        int n = t.size();
        vector<int> dp(n, 0);
        int resCenter = 0, resLen = 0;
        for (int i = 1; i < n; i++) {
            dp[i] =  mx > i ? min(mx - i, dp[2 * id - i]) : 1;
            while (t[i + dp[i]] == t[i - dp[i]])
                dp[i]++;
            if (mx < dp[i]+i) {
                mx = dp[i]+i;
                id = i;
            }
            if (dp[i] > resLen) {
                resLen = dp[i];
                resCenter = i;
            }
        }
        return s.substr((resCenter - resLen) / 2, resLen - 1);
    }
};

5.Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Thanks to @Freezen for additional test cases.

class Solution {
public:
    string shortestPalindrome(string s) {
        string r = s;
        reverse(r.begin(), r.end());
        string t = s + "#" + r;
        vector<int> next(t.size(), 0);
        for (int i = 1; i < t.size(); ++i) {
            int j = next[i - 1];
            while (j > 0 && t[i] != t[j]) j = next[j - 1];
            next[i] = (j += t[i] == t[j]);
        }
        return r.substr(0, s.size() - next.back()) + s;
    }
};

6.Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s ="aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

class Solution {
public:
    bool isPalindrome(string &s) {
        int n = s.length();
        if (n == 1) return true;
        int left = 0;
        int right = n - 1;
        while (left < right) {
            if (s[left] == s[right]) {
                left++;
                right--;
            } else {
                return false;
            }
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        if (s.empty()) return res;
        int n = s.length();
        for (int i = 1; i <= n; i++) {
            string substr = s.substr(0, i);
            if (isPalindrome(substr)) {
                vector<vector<string>> tmp = partition(s.substr(i));
                if (tmp.empty()) {
                    vector<string> v;
                    v.push_back(substr);
                    res.push_back(v);
                } else {
                    for (size_t j = 0; j < tmp.size(); j++) {
                        tmp[j].insert(tmp[j].begin(), substr);
                        res.push_back(tmp[j]);
                    }
                }
            }
        }
        return res;
    }
};