LeetCode —— Interleaving String

链接：http://leetcode.com/onlinejudge#question_97

原题：

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：

一开始用递归，思路很直观，但是小数据能过，大数据超时。

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;        return isInterleave(s1, 0, s2, 0, s3, 0);    }    private:    bool isInterleave(const string &s1, int i1,        const string &s2, int i2, const string &s3, int i3) {                if (i3 == s3.size())            return true;                if (i1 < s1.size() && s1[i1] == s3[i3]) {            if (isInterleave(s1, i1+1, s2, i2, s3, i3+1))                return true;        }                if (i2 < s2.size() && s2[i2] == s3[i3]) {            if (isInterleave(s1, i1, s2, i2+1, s3, i3+1))                return true;        }                return false;    }};

所以只能开个数组来记录状态

对于S3.size = n，检测能否由x长的S1和y长的S2交叉构成，这里x + y = n

只能由两种方式到达：

1） array[x-1][y] --> array[x][y], if S1[x] == S3[n]

2)    array[x][y-1] --> array[x][y], if S2[y] == S3[n]

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;        vector<vector<bool> > array;        for (int i=0; i<=s1.size(); i++) {            array.push_back(vector<bool>(s2.size()+1, false));        }        array[0][0] = true;        for (int n=1; n<=s3.size(); n++) {            for (int x=0, y=n; x<=n; x++, y--) {                if (x>=0 && x<=s1.size() && y>=0 && y<=s2.size()) {                    if ( (x-1 >= 0 && array[x-1][y] && s3[n-1]==s1[x-1]) ||                         (y-1 >= 0 && array[x][y-1] && s3[n-1]==s2[y-1]) )                        array[x][y] = true;                    else                        array[x][y] = false;                }                       }        }        return array[s1.size()][s2.size()];    }};

当然在空间上可以优化一下：

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;                if (s1.size() == 0 || s2.size() == 0) {            if (s1 == s3 || s2 == s3)                return true;            return false;        }                    vector<bool> vec(s2.size() + 1);        vector<vector<bool> > state;        state.push_back(vec);        state.push_back(vec);                int cur = 0;        int pre = 1;        state[cur][0] = true;        for (int i=0; i<s2.size(); i++)            if (s2[i] == s3[i])                state[cur][i+1] = true;            else                break;                for (int i=0; i<s1.size(); i++) {            cur = (cur + 1) % 2;            pre = (pre + 1) % 2;            if (state[pre][0] && s1[i] == s3[i])                state[cur][0] = true;            else                state[cur][0] = false;                        for (int j=0; j<s2.size(); j++) {                if ((s1[i] == s3[i+j+1] && state[pre][j+1]) || (s2[j] == s3[i+j+1] && state[cur][j]))                    state[cur][j+1] = true;                else                    state[cur][j+1] = false;            }        }                return state[cur].back();    }};