97. Interleaving String（dp）

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题目：

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.

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深搜：

超时

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        if(s1.length()+s2.length()!=s3.length())return false;        return dfs(s1,s2,s3,0,0,0);    }        private boolean dfs(String s1,String s2,String s3, int l1,int l2,int l3)    {    if(l3 == s3.length())return true;    if(s1.length()>l1 && s1.charAt(l1)==s3.charAt(l3))    {    if(dfs(s1,s2,s3,l1+1,l2,l3+1))    {    return true;    }    }    if(s2.length()>l2 && s2.charAt(l2)==s3.charAt(l3))    {    if(dfs(s1,s2,s3,l1,l2+1,l3+1))    {    return true;    }    }    return false;        }}

动态规划：

dp[i][j]表示s1的前i个和s2的前j个能构成s3的前i+j个

动态规划方程：dp[i][j] = dp[i - 1][j] && s2[i - 1] == s3[i + j - 1] || dp[i][j - 1] && s1[i][j - 1] == s3[i + j - 1]

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()+1;i++){if(s1.charAt(i-1)==s3.charAt(i-1) && dp[i-1][0]){dp[i][0]=true;}}for(int i=1;i<s2.length()+1;i++){if(s2.charAt(i-1) == s3.charAt(i-1)&& dp[0][i-1]){dp[0][i]=true;}}for(int i=1;i<s1.length()+1;i++){for(int j=1;j<s2.length()+1;j++){dp[i][j] = (s1.charAt(i-1)==s3.charAt(i+j-1) && dp[i-1][j])||(s2.charAt(j-1)==s3.charAt(i+j-1) && dp[i][j-1]);}}return dp[s1.length()][s2.length()];}}

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