Interleaving String

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

这是一道关于字符串操作的题目,要求是判断一个字符串能不能由两个字符串按照他们自己的顺序,每次挑取两个串中的一个字符来构造出来。

像这种判断能否按照某种规则来完成求是否或者某个量的题目,很容易会想到用动态规划来实现。

动态规划重点在于找到:维护量,递推式。维护量通过递推式递推,最后往往能得到想要的结果

先说说维护量,res[i][j]表示用s1的前i个字符和s2的前j个字符能不能按照规则表示出s3的前i+j个字符,如此最后结果就是res[s1.length()][s2.length()],判断是否为真即可。接下来就是递推式了,假设知道res[i][j]之前的所有历史信息,我们怎么得到res[i][j]。可以看出,其实只有两种方式来递推,一种是选取s1的字符作为s3新加进来的字符,另一种是选s2的字符作为新进字符。而要看看能不能选取,就是判断s1(s2)的第i(j)个字符是否与s3i+j个字符相等。如果可以选取并且对应的res[i-1][j](res[i][j-1])也为真,就说明s3i+j个字符可以被表示。这两种情况只要有一种成立,就说明res[i][j]为真,是一个或的关系。所以递推式可以表示成

res[i][j] = res[i-1][j] && s1.charAt(i-1)==s3.charAt(i+j-1) || 
            res[i][j-1] && s2.charAt(j-1)==s3.charAt(i+j-1)

Complexity

dp. O(MN) time & space.

Code

public class Solution {
    public boolean isInterleave_1(String s1, String s2, String s3) {
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if (l1 == 0) return s2.compareTo(s3) == 0;
        if (l2 == 0) return s1.compareTo(s3) == 0;
        if (l1 + l2 != l3) return false;
        boolean[][] dp = new boolean[l1+1][l2+1];
        dp[0][0] = true;
        for (int i = 1; i <= l1; ++i) {
            dp[i][0] = dp[i-1][0] && (s1.charAt(i-1) == s3.charAt(i-1));
        }
        for (int j = 1; j <= l2; ++j) {
            dp[0][j] = dp[0][j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
        }
        for (int i = 1; i <= l1; ++i) {
            for (int j = 1; j <= l2; ++j) {
                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i-1][j];
                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i][j] | dp[i][j-1];
            }
        }
        return dp[l1][l2];
    }
    
    public boolean isInterleave(String s1, String s2, String s3) {
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if (l1 == 0) return s2.compareTo(s3) == 0;
        if (l2 == 0) return s1.compareTo(s3) == 0;
        if (l1 + l2 != l3) return false;
        boolean[] dp = new boolean[l2+1];
        dp[0] = true;
        for (int j = 1; j <= l2; ++j) {
            dp[j] = dp[j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
        }
        for (int i = 1; i <= l1; ++i) {
            dp[0] = dp[0] && (s1.charAt(i-1) == s3.charAt(i-1));
            for (int j = 1; j <= l2; ++j) {
                boolean before = dp[j]; dp[j] = false;
                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[j] = before;
                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[j] = dp[j] | dp[j-1];
            }
        }
        return dp[l2];
    }
}