算法 中等 | 29. 交叉字符串

题目描述

给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。

样例1

输入:
"aabcc"
"dbbca"
"aadbbcbcac"
输出:
true

样例2

输入:
""
""
"1"
输出:
false

样例3

输入:
"aabcc"
"dbbca"
"aadbbbaccc"
输出:
false

解题思路

动态规划。
dp[i][j]代表由s1的前i个字母和s2的前j个字母是否能构成当前i+j个字母。
然后状态转移即可。（看第i+j+1个是否能被s1的第i+1个构成或被s2的第j+1个构成）

java题解

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

C++题解

class Solution {
public:
    /**
     * Determine whether s3 is formed by interleaving of s1 and s2.
     * @param s1, s2, s3: As description.
     * @return: true of false.
     */
    bool isInterleave(string s1, string s2, string s3) {
        if (s1.length() + s2.length() != s3.length()) {
            return false;
        }
        
        bool interleave[s1.length() + 1][s2.length() + 1];
        interleave[0][0] = true;
        for (int i = 1; i <= s1.length(); i++) {
            interleave[i][0] = interleave[i - 1][0] && s1[i - 1] == s3[i - 1];
        }
        for (int i = 1; i <= s2.length(); i++) {
            interleave[0][i] = interleave[0][i - 1] && s2[i - 1] == s3[i - 1];
        }
        
        for (int i = 1; i <= s1.length(); i++) {
            for (int j = 1; j <= s2.length(); j++) {
                interleave[i][j] = false;
                if (s1[i - 1] == s3[i + j - 1]) {
                    interleave[i][j] = interleave[i][j] || interleave[i - 1][j];
                }
                if (s2[j - 1] == s3[i + j - 1]) {
                    interleave[i][j] = interleave[i][j] || interleave[i][j - 1];
                }
            }
        }
        
        return interleave[s1.length()][s2.length()];
     }
};

python题解

class Solution:
    """
    @params s1, s2, s3: Three strings as description.
    @return: return True if s3 is formed by the interleaving of
             s1 and s2 or False if not.
    @hint: you can use [[True] * m for i in range (n)] to allocate a n*m matrix.
    """
    def isInterleave(self, s1, s2, s3):
        # write your code here
        if s1 is None or s2 is None or s3 is None:
            return False
        if len(s1) + len(s2) != len(s3):
            return False

        interleave = [[False] * (len(s2) + 1) for i in range(len(s1) + 1)]
        interleave[0][0] = True
        for i in range(len(s1)):
            interleave[i + 1][0] = s1[:i + 1] == s3[:i + 1]
        for i in range(len(s2)):
            interleave[0][i + 1] = s2[:i + 1] == s3[:i + 1]

        for i in range(len(s1)):
            for j in range(len(s2)):
                interleave[i + 1][j + 1] = False
                if s1[i] == s3[i + j + 1]:
                    interleave[i + 1][j + 1] = interleave[i][j + 1]
                if s2[j] == s3[i + j + 1]:
                    interleave[i + 1][j + 1] |= interleave[i + 1][j]
        return interleave[len(s1)][len(s2)]