【DP】交叉组合

题目描述

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.

comp[i][j]表示s1前i个字符跟s2前j个字符能否交叉组合成s3的前i+j个字符。

两种情况，如果comp[i][j-1]为true，观察s2中第j个字符跟s3中第i+j个字符是否相等。（因为做交叉的时候相对顺序不能改变）

如果comp[i-1][j]为true，观察s1中第i个字符跟s3中第i+j个字符是否相等。

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {                if(s1.length() + s2.length() != s3.length())            return false;                int lens1 = s1.length() + 1;        int lens2 = s2.length() + 1;        boolean [][] comp = new boolean [lens1][lens2];                comp[0][0] = true;                for(int i = 1; i < lens1; ++i){            comp[i][0] = comp[i - 1][0] && (s1.charAt(i - 1) == s3.charAt(i - 1));        }                for(int i = 1; i < lens2; ++i){            comp[0][i] = comp[0][i - 1] && (s2.charAt(i - 1) == s3.charAt(i - 1));        }                for(int i = 1; i < lens1; ++i){            for(int j = 1; j < lens2; ++j){                comp[i][j] = (comp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1))||                    (comp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));            }        }                return comp[lens1 - 1][lens2 - 1];    }}