判断一个字符串在至多删除k个字符后是否为回文串

转自： http://www.careercup.com/question?id=6287528252407808

问题描述：

A k-palindrome is a string which transforms into a palindrome on removing at most k characters.
Given a string S, and an interger K, print "YES" if S is a k-palindrome; otherwise print "NO".
Constraints:
S has at most 20,000 characters.
0<=k<=30
Sample Test Case#1:
Input - abxa 1
Output - YES
Sample Test Case#2:
Input - abdxa 1
Output – No

解答：懒得写了，下面这段通俗易懂，就先将就着看吧

The question asks if we can transform the given string S into its reverse deleting at most K letters.
We could modify the traditional Edit-Distance algorithm, considering only deletions, and check if this edit distance is <= K. There is a problem though. S can have length = 20,000 and the Edit-Distance algorithm takes O(N^2). Which is too slow.
(From here on, I'll assume you're familiar with the Edit-Distance algorithm and its DP matrix)
However, we can take advantage of K. We are only interested *if* manage to delete K letters. This means that any position more than K positions away from the main diagonal is useless because its edit distance must exceed those K deletions.
Since we are comparing the string with its reverse, we will do at most K deletions and K insertions (to make them equal). Thus, we need to check if the ModifiedEditDistance is <= 2*K
Here's the code:

   1:  int ModifiedEditDistance(const string& a, const string& b, int k) {

   2:      int i, j, n = a.size();

   3:      // for simplicity. we should use only a window of size 2*k+1 or 

   4:      // dp[2][MAX] and alternate rows. only need row i-1

   5:      int dp[MAX][MAX];

   6:      memset(dp, 0x3f, sizeof dp);    // init dp matrix to a value > 1.000.000.000

   7:      for (i = 0 ; i < n; i++)

   8:          dp[i][0] = dp[0][i] = i;

   9:   

  10:      for (i = 1; i <= n; i++) {

  11:          int from = max(1, i-k), to = min(i+k, n);

  12:          for (j = from; j <= to; j++) {

  13:              if (a[i-1] == b[j-1])            // same character

  14:                  dp[i][j] = dp[i-1][j-1];    

  15:              // note that we don't allow letter substitutions

  16:              

  17:              dp[i][j] = min(dp[i][j], 1 + dp[i][j-1]); // delete character j

  18:              dp[i][j] = min(dp[i][j], 1 + dp[i-1][j]); // insert character i

  19:          }

  20:      }

  21:      return dp[n][n];

  22:  }

  23:  cout << ModifiedEditDistance("abxa", "axba", 1) << endl;  // 2 <= 2*1 - YES

  24:  cout << ModifiedEditDistance("abdxa", "axdba", 1) << endl; // 4 > 2*1 - NO

  25:  cout << ModifiedEditDistance("abaxbabax", "xababxaba", 2) << endl; // 4 <= 2*2 - YES

We only process 2*K+1 columns per row. So this algorithm works in O(N*K) which is fast enough.