Posts Tagged 【dp】Interleaving String

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Interleaving String

Total Accepted: 29109 Total Submissions: 141918My Submissions

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.

/*感觉题目的interleaving给弄糊涂了s1=12345，s2=abcde，s3=123a45bcde===true反正就是从s1和s2中拿数据往s3上填就行Submission Result: Time Limit Exceeded*/public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        if(s1 == null || s2 == null || s3 == null) return false;        if(s1.length() + s2.length() != s3.length()) return false;        return isMatch(s1,0,s2,0,s3,0);    }    private boolean isMatch(String s1,int count1,String s2,int count2,String s3,int count3) {        if(count1 == s1.length() && count2 == s2.length()) return true;        if(count1 == s1.length()) return s2.substring(count2).equals(s3.substring(count3));        if(count2 == s2.length()) return s1.substring(count1).equals(s3.substring(count3));        if(s1.charAt(count1) == s3.charAt(count3) && s2.charAt(count2) == s3.charAt(count3))            return isMatch(s1,count1+1,s2,count2,s3,count3+1) || isMatch(s1,count1,s2,count2+1,s3,count3+1);        else if(s1.charAt(count1) == s3.charAt(count3))            return isMatch(s1,count1+1,s2,count2,s3,count3+1);        else if(s2.charAt(count2) == s3.charAt(count3))            return isMatch(s1,count1,s2,count2+1,s3,count3+1);        else return false;     }}

/*interleaving的关键就是，对s1,s2中取字符的顺序有要求，下一个字符必须是两个字串当前结尾的字符后面的第一个字符，不能跳过。dp的感觉就是优化，选中择优。从多个false、true中选中true也算，关键状态能转移就行。多个变量，多个条件，多维dp。二维[][]中>=1,>=1.dp[i][j] 代表取s1[0…i-1]中i个字符和s2[0…j-1]前j个字符，是不是能构成s3[0…k-1] (k==i+j)字符s3[k-1]只能来自s1[i-1]或者s2[j-1]。两种可能都满足，或者只有一个满足，或者如果都不满足就说明这种i,j组合不可能是interleaving了。*/public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        if(s1 == null || s2 == null || s3 == null) return false;        if(s1.length() + s2.length() != s3.length()) return false;        //"", "a", "",所有的+1        boolean dp[][] = new boolean[s1.length()+1][s2.length()+1];        dp[0][0] = true; //其他为false        for(int i = 0;i<s1.length()+1;i++)            for(int j = 0;j<s2.length()+1;j++) {                if(j>0 && s3.charAt(i+j-1) == s2.charAt(j-1) && dp[i][j-1])                    dp[i][j] = true;                else if(i>0 && s3.charAt(i-1+j) == s1.charAt(i-1))                    dp[i][j] = dp[i-1][j];            }        return dp[s1.length()][s2.length()];    }}

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