Interleaving String

原创于 2017-01-27 07:03:33 发布 · 91 阅读

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CC 4.0 BY-SA版权

本文介绍了一种用于判断字符串s3是否能由字符串s1和s2交错组成的有效算法。通过构建布尔矩阵，该方法能够在O(n^2)的时间复杂度内得出结论。文章通过实例演示了如何使用该算法进行判断。

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Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = “aabcc”,
s2 = “dbbca”,

When s3 = “aadbbcbcac”, return true.
When s3 = “aadbbbaccc”, return false.

import java.lang.Math; // headers MUST be above the first class
import java.util.*;

// one class needs to have a main() method
public class Solution {

   public boolean isInterleave(String s1, String s2, String s3) {

    if ((s1.length()+s2.length())!=s3.length()) return false;

    boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

    matrix[0][0] = true;

    for (int i = 1; i < matrix[0].length; i++){
        matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
    }

    System.out.println(Arrays.deepToString(matrix));

    for (int i = 1; i < matrix.length; i++){
        matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        for (int j = 1; j < matrix[0].length; j++){
            matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
                    || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
        }
    }

    System.out.println();
    System.out.println(Arrays.deepToString(matrix));
    return matrix[s2.length()][s1.length()];

    }

  public static void main(String[] args)
  {
    Solution test = new Solution();
    //int[] s1 = {10, 9, 2, 5, 3, 7, 101, 18,9,1,2};
    String s1 = "aabcc";
    String s2 = "dbbca";
    String s3 = "aadbbcbcac";
    String s4 = "aadbbbaccc";


    System.out.println(test.isInterleave(s1,s2,s3));
  }
}

[[true, true, true, false, false, false], [false, false, false, false, false, false], [false, false, false, false, false, false], [false, false, false, false, false, false], [false, false, false, false, false, false], [false, false, false, false, false, false]]

[[true, true, true, false, false, false], [false, false, true, true, false, false], [false, false, true, true, true, false], [false, false, true, false, true, true], [false, false, true, true, true, false], [false, false, false, false, true, true]]
true