动态规划-Minimum Insertion Steps to Make a String Palindrome
2020-01-05 11:52:40
问题描述:
问题求解:
好像多次碰到类似的lcs的变种题了,都是套上了回文的壳。这里再次记录一下。
其实本质就是裸的lcs,就出结果了。
public int minInsertions(String s) { StringBuffer sb = new StringBuffer(s); String b = sb.reverse().toString(); return s.length() - lcs(s, b); } public int lcs(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); int c[][] = new int[len1+1][len2+1]; for (int i = 0; i <= len1; i++) { for( int j = 0; j <= len2; j++) { if(i == 0 || j == 0) { c[i][j] = 0; } else if (str1.charAt(i-1) == str2.charAt(j-1)) { c[i][j] = c[i-1][j-1] + 1; } else { c[i][j] = Math.max(c[i - 1][j], c[i][j - 1]); } } } return c[len1][len2]; }