Palindrome Partitioning I,II[leetcode] DFS以及DP的解法

Palindrome Partitioning I,II[leetcode] DFS以及DP的解法

Palindrome Partitioning I

第一種方法是DFS，將所有可能的前綴找到，遞歸調用partition(剩余字符串)

復雜度為O(2^n)

代碼如下：

    vector> partition(string s) {
        vector> res;
        vector patition;
        if (s.size() == 0) return res;
        partition(s, patition, res);
        return res;
    }
    
    void partition(string s, vector& patition, vector>& res) {
        for (int i = 1; i < s.size(); i++)
        {
            if (isPalindrome(s.substr(0, i)))
            {
                patition.push_back(s.substr(0, i));
                partition(s.substr(i), patition, res);
                patition.pop_back();
            }
        }
        if (isPalindrome(s))
        {
            patition.push_back(s);
            res.push_back(patition);
            patition.pop_back();
        }
    }
    
    bool isPalindrome(string s)
    {
        int l = 0, r = s.size() - 1;
        while (l <= r)
        {
            if (s[l] != s[r]) return false;
            l++;
            r--;
        }
        return true;
    }

來自 https://oj.leetcode.com/discuss/9623/my-java-dp-only-solution-without-recursion-o-n-2

res(i)表示s[0...i-1]的所有分解方式

isPalin(i,j)表示s[i...j]是否為回文串

isPalin(i,j) = true if i==j or (s[i] == s[j] and isPalin(i + 1, j - 1)) or (s[i] == s[j] and i + 1 == j)

res(i) = res(j) + s[j...i-1] if isPalin(j, i-1)

    vector> partition(string s) {
        int size = s.size();
        vector>> res(size + 1);
        res[0].push_back(vector(0));
        vector> isPalin(size + 1, vector(size + 1, false));
        
        for (int i = 0; i < size; i++)
        {
            for (int j = 0; j <= i; j++)
            {
                if (i == j || s[i] == s[j] && (j + 1 == i || isPalin[j + 1][i - 1]))
                {
                    isPalin[j][i] = true;
                    for (int p = 0; p < res[j].size(); p++)
                    {
                        vector prefix = res[j][p];
                        prefix.push_back(s.substr(j, i - j + 1));
                        res[i + 1].push_back(prefix);
                    }
                }
            }
        }
        return res[size];
    }

Palindrome Partitioning II

按照Palindrome Partitioning I的DP的代碼，不難得到。

代碼如下：

   int minCut(string s) {
        int size = s.size();
        vector cut(size + 1, INT_MAX);
        vector> isPalin(size + 1, vector(size + 1, false));
        cut[0] = 0;
        
        for (int i = 0; i < size; i++)
        {
            for (int j = 0; j <= i; j++)
            {
                if (i == j || s[i] == s[j] && (j + 1 == i || isPalin[j + 1][i - 1]))
                {
                    isPalin[j][i] = true;
                    cut[i + 1] = min(cut[i + 1], cut[j] + 1);
                }
            }
        }
        return cut[size] - 1;
    }

第二種方法省去了isPalin數組

來自https://oj.leetcode.com/discuss/9476/solution-does-not-need-table-palindrome-right-uses-only-space

對於s中的位置i，分別尋找以i為中心的奇數長度回文子串s[i-j...i+j] 和偶數長度回文子串s[i-j+1...i+j]

從而有cut(i + j + 1) = min{cut(i - j) + 1, cut(i - j + 1) + 1}

代碼如下：

   int minCut(string s) {
        int n = s.size();
        vector cut(n+1, 0);  // number of cuts for the first k characters
        for (int i = 0; i <= n; i++) cut[i] = i-1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; i-j >= 0 && i+j < n && s[i-j]==s[i+j] ; j++) // odd length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j]);

            for (int j = 1; i-j+1 >= 0 && i+j < n && s[i-j+1] == s[i+j]; j++) // even length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j+1]);
        }
        return cut[n];
    }