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  • [leetcode]Scramble String

    三维DP,用备忘录方式写很简单。但是正如那天讨论时说的那样,使用备忘录法,可能会对复杂度和递推公式理解不够深刻。

    这道题目的递推公式很简单,但是复杂度的话。。。因为数组里的值每个都被计算仅为一次,所以应该是O(n^3)

    要注意的是,最后一维的是len+1

    public class Solution {
        public boolean isScramble(String s1, String s2) {
            char[] ca1 = s1.toCharArray();
            char[] ca2 = s2.toCharArray();
            if (s1.length() != s2.length()) return false;
            int len = s1.length();
            if (len == 0) return true;
            int[][][] mx = new int[len][len][len+1];
            for (int i = 0; i < len; i++)
                for (int j = 0; j < len; j++)
                    for (int k = 0; k < len+1; k++)
                        mx[i][j][k] = -1; // not visited;
            return isScramble(ca1, ca2, 0, 0, len, mx);
        }
        
        public boolean isScramble(char[] ca1, char[] ca2, int i1, int i2, int len, int[][][] mx)
        {
            if (len == 1) return ca1[i1] == ca2[i2];
            if (mx[i1][i2][len] != -1)
            {
                return mx[i1][i2][len] == 1;
            }
            for (int k = 1; k < len; k++)
            {
                if (isScramble(ca1, ca2, i1, i2, k, mx) && isScramble(ca1, ca2, i1+k, i2+k, len-k, mx))
                {
                    mx[i1][i2][len] = 1;
                    return true;
                }
                if (isScramble(ca1, ca2, i1, i2+len-k, k, mx) && isScramble(ca1, ca2, i1+k, i2, len-k, mx))
                {
                    mx[i1][i2][len] = 1;
                    return true;
                }
            }
            mx[i1][i2][len] = 0;
            return false;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/lautsie/p/3325408.html
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