LeetCode OJ-- Interleaving String **@

https://oj.leetcode.com/problems/interleaving-string/

刚开始用递归做，但是超时了

class Solution {
public:
    bool flag;
    
    bool isInterleave(string s1, string s2, string s3) {
        flag = false;
        if(s1.size() + s2.size() != s3.size())
            return flag;
            
        subIsInterleave(s1,s2,s3,0,0,0);
        return flag;
    }
    
    void subIsInterleave(string &s1, string &s2, string &s3, int p1, int p2, int p3)
    {
        if(p1 == s1.size() && p2 == s2.size() && p3 == s3.size())
        {
            flag = true;
            return;
        }
        if(!(p1 <= s1.size() && p2 <= s2.size() && p3 < s3.size()))
            return;
            
        if(s3[p3] != s1[p1] && s3[p3] != s2[p2])
        {
            return;
        }
        if(p1 < s1.size() && s3[p3] == s1[p1] && flag == false)
            subIsInterleave(s1,s2,s3,p1 + 1, p2,p3 + 1);
        if(p2 < s2.size() && s3[p3] == s2[p2] && flag == false)
            subIsInterleave(s1,s2,s3,p1,p2+1,p3 + 1);
    }
};

按照二维动态规划的思路，用两层for循环做

记 flag[i][j] 为 s1[0,i] s2[0,j] 匹配 s3[0,i+j] 则：

flag[i][j] = s1[i-1] == s3[i+j-1] && flag[i-1][j]  || s2[j-1] == s3[i+j-1] && flag[i][j-1];

代码如下：

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