Leetcode: Interleaving String

最新推荐文章于 2019-01-13 06:59:29 发布

LuckySeven7777

最新推荐文章于 2019-01-13 06:59:29 发布

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本文介绍如何使用动态规划解决字符串交织问题，通过给定的字符串s1和s2，判断是否存在一个字符串s3是由s1和s2交织而成。通过实例演示算法过程，并提供实现代码。

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Problem:

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.

Solution: DP

Just note that at the second time to assign value to result[i][j], should "||" the first result[i][j].

Code:

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1.length() + s2.length() != s3.length()) return false;
        boolean[][] result = new boolean[s1.length() + 1][s2.length() + 1];
        int count = 1;
        result[0][0] = true;
        while(count <= s1.length() && s1.charAt(count - 1) == s3.charAt(count - 1)) {
            result[count][0] = true; 
            count++;
        }
        count = 1;
        while (count <= s2.length() && s2.charAt(count - 1) == s3.charAt(count - 1)) {
            result[0][count] = true;
            count++;
        }
        for (int i = 1; i <= s1.length(); i++) {
            for (int j = 1; j <= s2.length(); j++) {
                if (s1.charAt(i - 1) == s3.charAt(i + j - 1)) {
                    result[i][j] = result[i - 1][j];
                    
                }
                if (s2.charAt(j - 1) == s3.charAt(i + j - 1)) {
                    result[i][j] = result[i][j - 1] || result[i][j];
                }
            }
        }
        return result[s1.length()][s2.length()];
    }
}