[LeetCode] 97. Interleaving String 交织相错的字符串

2021年09月15日 阅读数:1
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Given s1s2s3, find whether s3 is formed by the interleaving of s1and s2.html

Example 1:java

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:python

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

给定字符串s1, s2, s3,求s3是否能够由s1和s2交错造成。app

解法:DP动态规划,post

递推公式为:url

dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i - 1 + j]) || (dp[i][j - 1] && s2[j - 1] == s3[j - 1 + i]);spa

其中dp[i][j] 表示的是 s2 的前 i 个字符和 s1 的前 j 个字符是否匹配 s3 的前 i+j 个字符orm

Java:htm

public boolean isInterleave(String s1, String s2, String s3) {

    if ((s1.length()+s2.length())!=s3.length()) return false;

    boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

    matrix[0][0] = true;

    for (int i = 1; i < matrix[0].length; i++){
        matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        for (int j = 1; j < matrix[0].length; j++){
            matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
                    || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
        }
    }

    return matrix[s2.length()][s1.length()];

}

Python:blog

# O(m*n) space
def isInterleave1(self, s1, s2, s3):
    r, c, l= len(s1), len(s2), len(s3)
    if r+c != l:
        return False
    dp = [[True for _ in xrange(c+1)] for _ in xrange(r+1)]
    for i in xrange(1, r+1):
        dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
    for j in xrange(1, c+1):
        dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
    for i in xrange(1, r+1):
        for j in xrange(1, c+1):
            dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i-1+j]) or \
               (dp[i][j-1] and s2[j-1] == s3[i-1+j])
    return dp[-1][-1]

Python:

# O(2*n) space
def isInterleave2(self, s1, s2, s3):
    l1, l2, l3 = len(s1)+1, len(s2)+1, len(s3)+1
    if l1+l2 != l3+1:
        return False
    pre = [True for _ in xrange(l2)]
    for j in xrange(1, l2):
        pre[j] = pre[j-1] and s2[j-1] == s3[j-1]
    for i in xrange(1, l1):
        cur = [pre[0] and s1[i-1] == s3[i-1]] * l2
        for j in xrange(1, l2):
            cur[j] = (cur[j-1] and s2[j-1] == s3[i+j-1]) or \
                     (pre[j] and s1[i-1] == s3[i+j-1])
        pre = cur[:]
    return pre[-1]

Python:

# O(n) space
def isInterleave3(self, s1, s2, s3):
    r, c, l= len(s1), len(s2), len(s3)
    if r+c != l:
        return False
    dp = [True for _ in xrange(c+1)] 
    for j in xrange(1, c+1):
        dp[j] = dp[j-1] and s2[j-1] == s3[j-1]
    for i in xrange(1, r+1):
        dp[0] = (dp[0] and s1[i-1] == s3[i-1])
        for j in xrange(1, c+1):
            dp[j] = (dp[j] and s1[i-1] == s3[i-1+j]) or (dp[j-1] and s2[j-1] == s3[i-1+j])
    return dp[-1]

Python:

# DFS 
def isInterleave4(self, s1, s2, s3):
    r, c, l= len(s1), len(s2), len(s3)
    if r+c != l:
        return False
    stack, visited = [(0, 0)], set((0, 0))
    while stack:
        x, y = stack.pop()
        if x+y == l:
            return True
        if x+1 <= r and s1[x] == s3[x+y] and (x+1, y) not in visited:
            stack.append((x+1, y)); visited.add((x+1, y))
        if y+1 <= c and s2[y] == s3[x+y] and (x, y+1) not in visited:
            stack.append((x, y+1)); visited.add((x, y+1))
    return False

Python:  

# BFS 
def isInterleave(self, s1, s2, s3):
    r, c, l= len(s1), len(s2), len(s3)
    if r+c != l:
        return False
    queue, visited = [(0, 0)], set((0, 0))
    while queue:
        x, y = queue.pop(0)
        if x+y == l:
            return True
        if x+1 <= r and s1[x] == s3[x+y] and (x+1, y) not in visited:
            queue.append((x+1, y)); visited.add((x+1, y))
        if y+1 <= c and s2[y] == s3[x+y] and (x, y+1) not in visited:
            queue.append((x, y+1)); visited.add((x, y+1))
    return False

 

Python:

# Time:  O(m * n)
# Space: O(m + n)
class Solution(object):
    # @return a boolean
    def isInterleave(self, s1, s2, s3):
        if len(s1) + len(s2) != len(s3):
            return False
        if len(s1) > len(s2):
            return self.isInterleave(s2, s1, s3)
        match = [False for i in xrange(len(s1) + 1)]
        match[0] = True
        for i in xrange(1, len(s1) + 1):
            match[i] = match[i -1] and s1[i - 1] == s3[i - 1]
        for j in xrange(1, len(s2) + 1):
            match[0] = match[0] and s2[j - 1] == s3[j - 1]
            for i in xrange(1, len(s1) + 1):
                match[i] = (match[i - 1] and s1[i - 1] == s3[i + j - 1]) \
                                       or (match[i] and s2[j - 1] == s3[i + j - 1])
        return match[-1]

Python:

# Time:  O(m * n)
# Space: O(m * n)
# Dynamic Programming
class Solution2(object):
    # @return a boolean
    def isInterleave(self, s1, s2, s3):
        if len(s1) + len(s2) != len(s3):
            return False
        match = [[False for i in xrange(len(s2) + 1)] for j in xrange(len(s1) + 1)]
        match[0][0] = True
        for i in xrange(1, len(s1) + 1):
            match[i][0] = match[i - 1][0] and s1[i - 1] == s3[i - 1]
        for j in xrange(1, len(s2) + 1):
            match[0][j] = match[0][j - 1] and s2[j - 1] == s3[j - 1]
        for i in xrange(1, len(s1) + 1):
            for j in xrange(1, len(s2) + 1):
                match[i][j] = (match[i - 1][j] and s1[i - 1] == s3[i + j - 1]) \
                                       or (match[i][j - 1] and s2[j - 1] == s3[i + j - 1])
        return match[-1][-1]

Python:  

# Time:  O(m * n)
# Space: O(m * n)
# Recursive + Hash
class Solution3(object):
    # @return a boolean
    def isInterleave(self, s1, s2, s3):
        self.match = {}
        if len(s1) + len(s2) != len(s3):
            return False
        return self.isInterleaveRecu(s1, s2, s3, 0, 0, 0)

    def isInterleaveRecu(self, s1, s2, s3, a, b, c):
        if repr([a, b]) in self.match.keys():
            return self.match[repr([a, b])]

        if c == len(s3):
            return True

        result = False
        if a < len(s1) and s1[a] == s3[c]:
            result = result or self.isInterleaveRecu(s1, s2, s3, a + 1, b, c + 1)
        if b < len(s2) and s2[b] == s3[c]:
            result = result or self.isInterleaveRecu(s1, s2, s3, a, b + 1, c + 1)

        self.match[repr([a, b])] = result

        return result

C++:

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

  

 

 

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