[leetcode刷题系列]Interleaving String

经典dp题，O(n^2)的状态，O(1)的转移。

const int MAXN = 1000 + 10;string s1, s2, s3;int dp[MAXN][MAXN];int is(int i, int j){    int &ret = dp[i][j];    if(i == 0 && j == 0)        return 1;    if(ret != -1)        return ret;    if(i > 0)        if(s1[i - 1] == s3[i + j - 1])            if(is(i - 1, j))                return ret = 1;    if(j > 0)        if(s2[j - 1] == s3[i + j - 1])            if(is(i, j - 1))                return ret = 1;    return ret = 0;}class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        // Start typing your C/C++ solution below        // DO NOT write int main() function            if(s1.size() + s2.size() != s3.size())            return false;        ::s1 = s1;        ::s2 = s2;        ::s3 = s3;        memset(dp, 0xff, sizeof(dp));        return is(s1.size(), s2.size());    }};