97.Interleaving-String

97. Interleaving String

题目地址

https://www.jiuzhang.com/solutions/palindrome-partitioning-ii/

题目描述

Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1 and s2.

思路

state: f[i][j] 表示s1的前i个字符和s2的前j个字符能否交替组成s3的前i+j个字符

function: f[i][j] = 
f[i - 1][j] && (s1[i - 1] == s3[i + j - 1])
f[i][j - 1] && (s2[j - 1] == s3[i + j - 1])

initialize: 
f[i][0] = (s1[0..i - 1] == s3[0..i - 1])
f[0][j] = (s2[0..j - 1] == s3[o..j - 1])

代码

Approach #1 Dynamic Programming

public class Solution {
  public boolean isInterleave(String s1, String s2, String s3) {
    if (s1.length() + s2.length() != s3.length()) {
        return false;
    }

    boolean[][] f = new boolean[s1.length() + 1][s2.length() + 1];
    f[0][0] = true;

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

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

    return f[s1.length()][s2.length()];
  }
}