首页 > 代码库 > 【LeetCode】Interleaving String

【LeetCode】Interleaving String

Interleaving String

Given s1s2s3, 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.

 

可以用递归做,每匹配s1或者s2中任意一个就递归下去。但是会超时。

因此考虑用动态规划做。

s1, s2只有两个字符串,因此可以展平为一个二维地图,判断是否能从左上角走到右下角。

当s1到达第i个元素,s2到达第j个元素:

地图上往右一步就是s2[j-1]匹配s3[i+j-1]。

地图上往下一步就是s1[i-1]匹配s3[i+j-1]。

示例:s1="aa",s2="ab",s3="aaba"。标1的为可行。最终返回右下角。

     0  a  b

0   1  1  0

a   1  1  1

a   1  0  1

 

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        if(s1.size()+s2.size() != s3.size())            return false;        //s1.size()+1 by s2.size()+1 matrix representing path        vector<vector<bool> > path(s1.size()+1, vector<bool>(s2.size()+1, false));        path[0][0] = true;        for(int i = 0; i < s1.size()+1; i ++)        {//row            for(int j = 0; j < s2.size()+1; j ++)            {//col                if(i == 0 && j == 0)                //start                    ;                else if(i == 0)                //go by s2                    path[i][j] = path[i][j-1] && (s2[j-1] == s3[j-1]);                else if(j == 0)                //go by s1                    path[i][j] = path[i-1][j] && (s1[i-1] == s3[i-1]);                else                    path[i][j] = (path[i][j-1] && (s2[j-1] == s3[i+j-1])) || (path[i-1][j] && (s1[i-1] == s3[i+j-1]));            }        }        return path[s1.size()][s2.size()];    }};

【LeetCode】Interleaving String