Interleaving String

Change-Id: I1990bbaf997953af5bf6de3e18c8a216972517ca

Author: applewjg

InterleavingString.java

@@ -0,0 +1,62 @@
+/*
+ Author:     King, [email protected]
+ Date:       Jan 3, 2015
+ Problem:    Interleaving String
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/interleaving-string/
+ Notes:
+ 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.
+
+ Solution: 1. dp. O(MN) time & space. 'dp[i][j] == true' means that there is at least one way to construct 
+              the string s3[0...i+j-1] by interleaving s1[0...j-1] and s2[0...i-1].
+ */
+
+public class Solution {
+    public boolean isInterleave_1(String s1, String s2, String s3) {
+        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
+        if (l1 == 0) return s2.compareTo(s3) == 0;
+        if (l2 == 0) return s1.compareTo(s3) == 0;
+        if (l1 + l2 != l3) return false;
+        boolean[][] dp = new boolean[l1+1][l2+1];
+        dp[0][0] = true;
+        for (int i = 1; i <= l1; ++i) {
+            dp[i][0] = dp[i-1][0] && (s1.charAt(i-1) == s3.charAt(i-1));
+        }
+        for (int j = 1; j <= l2; ++j) {
+            dp[0][j] = dp[0][j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
+        }
+        for (int i = 1; i <= l1; ++i) {
+            for (int j = 1; j <= l2; ++j) {
+                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i-1][j];
+                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i][j] | dp[i][j-1];
+            }
+        }
+        return dp[l1][l2];
+    }
+    public boolean isInterleave(String s1, String s2, String s3) {
+        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
+        if (l1 == 0) return s2.compareTo(s3) == 0;
+        if (l2 == 0) return s1.compareTo(s3) == 0;
+        if (l1 + l2 != l3) return false;
+        boolean[] dp = new boolean[l2+1];
+        dp[0] = true;
+        for (int j = 1; j <= l2; ++j) {
+            dp[j] = dp[j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
+        }
+        for (int i = 1; i <= l1; ++i) {
+            dp[0] = dp[0] && (s1.charAt(i-1) == s3.charAt(i-1));
+            for (int j = 1; j <= l2; ++j) {
+                boolean before = dp[j]; dp[j] = false;
+                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[j] = before;
+                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[j] = dp[j] | dp[j-1];
+            }
+        }
+        return dp[l2];
+    }
+}