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.

动态规划

思路

state: dp[i][j] s1取前i个字符 s2取前j个字符 s3取前i+1字符 是否能匹配
function:  如果最后一位(i+j位)与s1的最后一位(i位)相等 dp[i][j] = dp[i-1][j]
与s2最后一位相等则dp[i][j] = dp[i][j-1]
initial: dp[0][0] = true dp[i][0] dp[0][j]看s1 s2 s3比较
返回: dp[s1.length][s2.length]

复杂度

时间O(mn) 空间O(mn)

代码

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

一维动态规划

复杂度

时间O(m*n) 空间O(min(m,n))

代码

public boolean isInterleave(String s1, String s2, String s3) {
    if (s1.length() + s2.length() != s3.length()) {
        return false;
    }
    String maxWord = s1.length() > s2.length() ? s1 : s2;
    String minWord = s1.length() > s2.length() ? s2 : s1;
    boolean[]dp = new boolean[minWord.length() + 1];
    dp[0] = true;
   
    for (int i = 0; i < minWord.length(); i++) {
        if (minWord.charAt(i) == s3.charAt(i) && dp[i] == true) {
            dp[i + 1] = true;
        }
    }
    for (int i = 1; i <= maxWord.length(); i++) {
        dp[0] = dp[0] && maxWord.charAt(i - 1) == s3.charAt(i - 1); 
        for (int j = 1; j <= minWord.length(); j++) {
            dp[j] = s3.charAt(i + j - 1) == maxWord.charAt(i - 1) && dp[j] == true || s3.charAt(i + j - 1) == minWord.charAt(j - 1) && dp[j - 1] == true;
            //此处不能再用if判断, 因为dp[j]已经有值 必须要赋值true/false
        }
    }
    
    return dp[minWord.length()];
}

lpy1990
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