LeetCode--Interleaving String

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.

思路：动态规划。
这种字符串交错组合的题目是典型的dp问题，状态dp[l1][l2]表示长度l1的s1和长度l2的s2能否通过交错组合找到s3，初始化状态dp[0][0],dp[0][i],dp[i][0]后，状态方程为dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1]).

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        int l1=s1.size(),l2=s2.size(),l3=s3.size();        if(l1+l2!=l3) return false;        bool dp[l1+1][l2+1];        dp[0][0]=true;        for(int i=1;i<=l1;i++){            dp[i][0]=(s1[i-1]==s3[i-1]&&dp[i-1][0]);        }        for(int i=1;i<=l2;i++){            dp[0][i]=(s2[i-1]==s3[i-1]&&dp[0][i-1]);        }        for(int i=1;i<=l1;i++){            for(int j=1;j<=l2;j++){                dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1]);            }        }        return dp[l1][l2];    }};