97. Interleaving String

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dfs:
题意就是s1和s2是否可以交叉变成s3，一开始我使用dfs，但是超时了
然后剪枝，加入一个标记数组，如果dfs过的就不进行dfs，然后就ac了

class Solution {public:    int mapp[500][500];    int mark = 0;    void pipei(string s1, string s2, string s3, int a, int b, int c){        if(mark) return ;        mapp[a][b] = 1;        if(a + 1 == s1.length() && b == s2.length() && c + 1 == s3.length() && s1[a] == s3[c]){            mark = 1;            return ;        }        else if(a == s1.length() && b + 1 == s2.length() && c + 1 == s3.length() && s2[b] == s3[c]){            mark = 1;            return ;        }        else if(a < s1.length() && b < s2.length()){            if(s1[a] == s3[c] && !mapp[a + 1][b])                pipei(s1, s2, s3, a + 1, b, c + 1);            if(s2[b] == s3[c] && !mapp[a][b + 1])                pipei(s1, s2, s3, a, b + 1, c + 1);        }        else if(a < s1.length()){            if(s1[a] == s3[c] && !mapp[a + 1][b])                pipei(s1, s2, s3, a + 1, b, c + 1);        }        else if(b < s2.length()){            if(s2[b] == s3[c] && !mapp[a][b + 1])                pipei(s1, s2, s3, a, b + 1, c + 1);        }        return ;    }    bool isInterleave(string s1, string s2, string s3) {        if(s1.length() + s2.length() != s3.length()) return false;        if(s1.length() == 0 && s2.length() == 0 && s3.length() == 0) return true;        memset(mapp, 0, sizeof(mapp));        pipei(s1, s2, s3, 0, 0, 0);        if(mark) return true;        else return false;    }};

dp:
其实这题应该用dp，mapp (i,j)代表到s1 i 和s2 j都可以匹配，然后方程是
if(mapp[i - 1][j] && s1[i - 1] == s3[i + j - 1])
mapp[i][j] = true;
if(mapp[i][j - 1] && s2[j - 1] == s3[i + j - 1])
mapp[i][j] = true;
十分好理解

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        if(s1.length() + s2.length() != s3.length()) return false;        if(s1.length() == 0 && s2.length() == 0 && s3.length() == 0) return true;        bool mapp[s1.length() + 1][s2.length() + 1];        memset(mapp, false, sizeof(mapp));        for(int i = 0; i <= s1.length(); ++ i)            for(int j = 0; j <= s2.length(); ++ j){                if(i == 0 && j == 0)                    mapp[i][j] = true;                else if(i == 0){                    if(mapp[0][j - 1] && s2[j - 1] == s3[j - 1])                        mapp[0][j] = true;                }                else if(j == 0){                    if(mapp[i - 1][0] && s1[i - 1] == s3[i - 1])                        mapp[i][0] = true;                }                else{                    if(mapp[i - 1][j] && s1[i - 1] == s3[i + j - 1])                        mapp[i][j] = true;                    if(mapp[i][j - 1] && s2[j - 1] == s3[i + j - 1])                        mapp[i][j] = true;                }            }        return mapp[s1.length()][s2.length()];    }};

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