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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.
State:f[i][j] 表示s1的前i 个字符 和 s2的前j 个字符能组成s3的前 i + j 个字符
Function: if (((s1.charAt(i - 1) == s3.charAt(i + j - 1) && interleave[i - 1][j])) || (s2.charAt(j - 1) == s3.charAt(i + j - 1) && interleave[i][j - 1])) {
interleave[i][j] = true;
Initializtion:f[0][0] = true f[i][0] i = (1 ~ s1.length() ) f[0][j] j = (1 ~ s2.length())
Answer:f[s1.length()][s2.length()]
1 public class Solution { 2 public boolean isInterleave(String s1, String s2, String s3) { 3 if (s1.length() + s2.length() != s3.length()) { 4 return false; 5 } 6 7 boolean[][] interleave = new boolean[s1.length() + 1][s2.length() + 1]; 8 interleave[0][0] = true; 9 for (int i = 1; i <= s1.length(); i++) { 10 if (s1.charAt(i - 1) == s3.charAt(i - 1) && interleave[i - 1][0]) { 11 interleave[i][0] = true; 12 } 13 } 14 for (int i = 1; i <= s2.length(); i++) { 15 if (s2.charAt(i - 1) == s3.charAt(i - 1) && interleave[0][i - 1]) { 16 interleave[0][i] = true; 17 } 18 } 19 20 for (int i = 1; i <= s1.length(); i++) { 21 for (int j = 1; j <= s2.length(); j++) { 22 if (((s1.charAt(i - 1) == s3.charAt(i + j - 1) && interleave[i - 1][j])) || (s2.charAt(j - 1) == s3.charAt(i + j - 1) && interleave[i][j - 1])) { 23 interleave[i][j] = true; 24 } 25 } 26 } 27 return interleave[s1.length()][s2.length()]; 28 } 29 }
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原文地址：http://www.cnblogs.com/FLAGyuri/p/5405628.html
