LeetCode:Interleaving String

题目描述：

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;
else
break;
}

//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;
else
break;
}

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];
}
};