leetcode Interleaving String

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.

使用动态规划，思路如下：

这是一个二维的动态规划，如s1长len1，s2长len2，建立一个(len1+1)*(len2+1)的bool数组，元素均初始化false

F[i][j]代表 s3[i+j-1] 与 s1[0]~s1[i-1]、s2[0]~s2[j] 或 s1[0]~s1[i]、s2[0]~s2[j-1] 是否满足题中条件，判断其为true还是false的条件就是 F[i-1][j] 为 true 且 s3[i-1+j] == s1[i-1]，或者 F[i][j-1] 为 true 且 s3[i+j-1] == s2[j-1]。

true    ; (i=0 && j=0)

F[i][j] = {

F[i-1][j] && (s1[i-1] == s3[i-1+j]) || F[i][j-1] && (s2[j-1] == s3[i+j-1]) ;(i>0 ||　j>0)

代码如下：

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();if(len3 != len1+len2)return false;vector<vector<bool> >Tmp(len1+1, vector<bool>(len2+1, false));//初始化Tmp[0][0]为truebool tmp1 ,tmp2;for(int i=0; i<=len1; ++i){for(int j=0; j<=len2; ++j){tmp1 = false;tmp2 = false;if(i == 0 && j == 0){Tmp[i][j] = true;continue;}if(i > 0)tmp1 = Tmp[i-1][j] && (s1[i-1] == s3[i-1+j]);if(j > 0)tmp2 = Tmp[i][j-1] && (s2[j-1] == s3[i+j-1]);Tmp[i][j] = tmp1 || tmp2;}}return Tmp[len1][len2];    }};