动态规划-Minimum Insertion Steps to Make a String Palindrome

2020-01-05 11:52:40

问题描述:

动态规划-Minimum Insertion Steps to Make a String Palindrome

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];  
    }

  

 

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