[Lintcode] Interleaving String 交叉字符串

最新推荐文章于 2019-06-03 09:21:58 发布

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最新推荐文章于 2019-06-03 09:21:58 发布

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分类专栏：算法文章标签： lintcode

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本文介绍了一种判断字符串s3是否由s1和s2交错组成的高效算法，采用动态规划方法解决该问题，并提供了具体实现代码。

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Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1and s2.

Have you met this question in a real interview?

Yes

Example

For s1 = "aabcc", s2 = "dbbca"

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

要求时间复杂度小于O（N×N）。分析：关于字符串比较的问题，可以考虑使用DP求解。这里可以使用二维DP。

res[i][j] 代表s1中前i个字符与s2中前j个字母是否可以交叉匹配成为s2中前i+j个字符。

公式为：

if s3[i + j] = s1[i] = s2[j]. res[i][j] = res[i - 1][j] || res[i][j - 1]
if s3[i + j] = s1[i] res[i][j] = res[i - 1][j]
if s3[i + j] = s2[j] res[i][j] = res[i][j - 1]
else s3[i + j] = false;

初始条件是res[0][0] = true. 即s1前0个字符加s2前0个字符，可以拼凑成s3前0个字符。

public class Solution {
    /**
     * Determine whether s3 is formed by interleaving of s1 and s2.
     * @param s1, s2, s3: As description.
     * @return: true or false.
     */
    public boolean isInterleave(String s1, String s2, String s3) {

        if(s1.length() + s2.length() != s3.length()) return false;
        
        boolean[][] res = new boolean[s1.length() + 1][s2.length() + 1];
        res[0][0] = true;
        for(int i = 1; i <= s1.length(); i++) 
            res[i][0] = res[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1);
        for(int j = 1; j <= s2.length(); j++) 
            res[0][j] = res[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
        for(int i = 1; i <= s1.length(); i++) {
            for(int j = 1; j <= s2.length(); j++) {
                if(s3.charAt(i + j - 1) == s1.charAt(i - 1) &&
                s3.charAt(i + j - 1) == s2.charAt(j - 1)) {
                    res[i][j] = res[i - 1][j] || res[i][j - 1];
                } else if (s3.charAt(i + j - 1) == s1.charAt(i - 1)) {
                    res[i][j] = res[i - 1][j];
                } else if (s3.charAt(i + j - 1) == s2.charAt(j - 1)) {
                    res[i][j] = res[i][j - 1];
                } else 
                    res[i][j] = false;
            }
        }
        
        return res[s1.length()][s2.length()];
    }
}