题目描述：
Givens1,s2,s3, find whethers3is formed by the interleaving ofs1ands2.
For example,Given:s1="aabcc",s2="dbbca",
Whens3="aadbbcbcac", return true.Whens3="aadbbbaccc", return false.
思路分析：设s2的长度为length2，，s1的长度为length1。开一个(length2+1)*(length1+1)的bool数组flag。flag[i][j]表示字符串s1+i和字符串s2+j能否构造s3+i+j。首先初始化二维数组的最后一列和最后一行，根据状态转移方程来求得flag[0][0]，即得结果。
代码：
class Solution{public:bool isInterleave(string s1,string s2,string s3){int index1 = 0,index2 = 0,index3 = 0;int length1 = s1.length();int length2 = s2.length();int length3 = s3.length();if((length1 + length2) != length3)return false;if(length1 == 0)return s2 == s3;if(length2 == 0)return s1 == s3;//生成一个(length2+1)*(length1+1)的bool数组vector<vector<bool>> flag;for(int i = 0;i < (length2 + 1);i++){vector<bool> temp(length1+1,false);flag.push_back(temp);}//初始化最后一列for(int i = length2 – 1;i >= 0;i–){if(s2[i] == s3[length1 + i])flag[i][length1] = true;elsebreak;}//初始化最后一行for(int i = length1 – 1;i >= 0;i–){if(s1[i] == s3[length2 + i])flag[length2][i] = true;elsebreak;}for(int i = length2 – 1;i >= 0;i–)for(int j = length1 – 1;j >= 0;j–){if(s1[j] == s3[i + j])if(flag[i][j+1]){flag[i][j] = true;continue;}if(s2[i] == s3[i + j])if(flag[i+1][j]){flag[i][j] = true;continue;}}return flag[0][0];}};
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