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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.
题解 类似题目还有Scramble String， 一看就是boolean[][]... dp的咯，这种问题最关键的就是理解dp的概念
引用ref(http://www.cnblogs.com/springfor/p/3896159.html)里的话“动态规划数组是dp[i][j]，表示：s1取前i位，s2取前j位，是否能组成s3的前i+j位。”
这样code思路就很好理解咯
public class Solution { 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=1;i<=s1.length();i++){ dp[i][0] = dp[i-1][0]&&(s1.charAt(i-1)==s3.charAt(i-1)); } for(int j=1;j<=s2.length();j++){ dp[0][j] = dp[0][j-1]&&(s2.charAt(j-1)==s3.charAt(j-1)); } for(int i=1;i<=s1.length();i++){ for(int j=1;j<=s2.length();j++){ dp[i][j] =(dp[i-1][j] && (s1.charAt(i-1)==s3.charAt(i+j-1)))||(dp[i][j-1] && (s2.charAt(j-1)==s3.charAt(i+j-1))); } } return dp[s1.length()][s2.length()]; } }
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原文地址：http://www.cnblogs.com/jiajiaxingxing/p/4564988.html
