[LeetCode] Interleaving String

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本文介绍了一种判断字符串s3是否由s1和s2交错组成的算法。通过递归方法解决小规模问题，并采用动态规划解决大规模问题。给出了C++及Java实现代码。

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.

First we can consider this as a recursive problem.
If s3[i1+i2+1] == s1[i1+1], search(i1+1,i2)
If s3[i1+i2+1] == s2[i2+1], search(i1,i2+1)
until end. However, this can only pass the small test cases, but failed on the large test cases.

So we need to think of the dynamic programming (DP), which usually have much less complexity. In this problem, a 2D DP is more suitable. As usual, the typical way of solving dp is to find the state, and the optimal function. Here, the state can be considered as: A[i][j], which means S3[i+j] can be formed by S1[i] and S2[j] (for simplicity here string starts from 1, in the code we need to deal with that string starts from 0).

So, we have the optimal function:
A[i][j] = (s3[i+j]==s1[i] && match[i-1][j]) || (s3[i+j] ==s2[j] && match[i][j-1])

c++

bool isInterleave(string s1, string s2, string s3) {
         if(s1.size()+s2.size() != s3.size()) return false;
    bool **dp = new bool *[s1.size()+1];
    for(int i=0; i<s1.size()+1; i++){
        dp[i] = new bool [s2.size()+1];
    }
    dp[0][0] = true;
    for(int i=1; i<s1.size()+1; i++){
        if(s1[i-1]==s3[i-1] && dp[i-1][0])
            dp[i][0] = true;
    }
    for(int j=1; j<s2.size()+1; j++){
        if(s2[j-1]==s3[j-1] && dp[0][j-1])
            dp[0][j] = true;
    }
    for(int i=1; i<s1.size()+1; i++){
        for(int j=1; j<s2.size()+1; j++){
            if(s1[i-1]==s3[i+j-1] && dp[i-1][j])
                dp[i][j] = true;
            if(s2[j-1]==s3[i+j-1] && dp[i][j-1])
                dp[i][j] = true;
        }
    }
    return dp[s1.size()][s2.size()];
    }

java

public boolean isInterleave(String s1, String s2, String s3) {
        int n1 = s1.length();
        int n2 = s2.length();
        if(n1+n2!=s3.length()) return false;
        boolean [][]flag = new boolean [n1+1][n2+1];
        for(int i=0;i<=n1;i++){
        	for(int j=0;j<=n2;j++){
        		flag[i][j] = false;
        	}
        }
        flag[0][0] = true;
        for(int i=1;i<=n1;i++){
        	if(s1.charAt(i-1)==s3.charAt(i-1)&& flag[i-1][0]) flag[i][0] = true;
        }
        for(int j=1;j<=n2;j++){
        	if(s2.charAt(j-1)==s3.charAt(j-1)&& flag[0][j-1]) flag[0][j] =true;
        }
        for(int i=1;i<=n1;i++){
        	for(int j=1;j<=n2;j++){
        		if((flag[i-1][j] && s1.charAt(i-1)==s3.charAt(i-1+j))||
        				(flag[i][j-1]&& s2.charAt(j-1)==s3.charAt(i+j-1))){
        			flag[i][j] = true;
        		}
        	}
        }
        return flag[n1][n2];
    }