交错字符串
给定三个字符串 s1、s2、s3,请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。
两个字符串 s 和 t 交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空 子字符串:
s = s1 + s2 + … + sn
t = t1 + t2 + … + tm
|n - m| <= 1
交错 是 s1 + t1 + s2 + t2 + s3 + t3 + … 或者 t1 + s1 + t2 + s2 + t3 + s3 + …
提示:a + b 意味着字符串 a 和 b 连接。示例 1:
输入:s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbcbcac”
输出:true
示例 2:输入:s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”
输出:false
示例 3:输入:s1 = “”, s2 = “”, s3 = “”
输出:true提示:
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、和 s3 都由小写英文字母组成
- 思路是两个
- 一是回溯,但是由于没有剪枝,所以最后一个例子过不了
- 二是dp,dp代表取前i个s1的字母和前j个s2的字母是否能匹配前i+j个s3的字母,转换表达式为:
dp[i][j] = (dp[i][j - 1] && st3[i + j - 1] == st2[j - 1]) || (dp[i - 1][j] && st3[i + j - 1] == st1[i - 1]);
在init完毕第一行和第一列之后,dp[1][1]到dp[len1][len2]
的数由这个式子完成
class Solution
{
public:
int len1, len2, len3;
string st1, st2, st3;
bool isInterleave(string s1, string s2, string s3)
{
if (s1 == "" && s2 == "" && s3 == "")
return true;
len1 = s1.size();
len2 = s2.size();
len3 = s3.size();
st1 = s1;
st2 = s2;
st3 = s3;
if (len1 + len2 != len3)
return false;
else if (len1 == 0)
return st2.compare(st3) == 0;
else if (len2 == 0)
return st1.compare(st3) == 0;
// return judge(0, 0, 0);
return judge_v2();
}
bool judge_v2()
{
// bool dp[len1][len2];
vector<vector<bool>> dp(len1 + 1, vector<bool>(len2 + 1, false));
//dp初始化
dp[0][0] = true;
int jud = true;
for (int i = 1; i <= len1; i++)
{
if (st1[i - 1] != st3[i - 1])
jud = false;
dp[i][0] = jud;
}
jud = true;
for (int i = 1; i <= len2; i++)
{
if (st2[i - 1] != st3[i - 1])
jud = false;
dp[0][i] = jud;
}
for (int i = 1; i < len1 + 1; i++)
{
for (int j = 1; j < len2 + 1; j++)
{
dp[i][j] = (dp[i][j - 1] && st3[i + j - 1] == st2[j - 1]) || (dp[i - 1][j] && st3[i + j - 1] == st1[i - 1]);
}
}
return dp[len1][len2];
}
bool judge(int index1, int index2, int index3)
{
//何时返回true?
//有一边比较完毕了
if (index1 == len1)
return st2.compare(index2, len2 - index2, st3, index3, len3 - index3) == 0;
else if (index2 == len2)
return st1.compare(index1, len1 - index1, st3, index3, len3 - index3) == 0;
//何时直接返回false? st1[index1]!=st3[index3] && st2[index2]!=st3[index3]
if (st1[index1] != st3[index3] && st2[index2] != st3[index3])
return false;
//单边相等
else if (st1[index1] == st3[index3] && st2[index2] != st3[index3])
{
return judge(index1 + 1, index2, index3 + 1);
}
else if (st1[index1] != st3[index3] && st2[index2] == st3[index3])
{
return judge(index1, index2 + 1, index3 + 1);
}
//何时分支?s1[index1]==s2[index2] && s1[index1]==s3[index3]
else if (st1[index1] == st2[index2] && st1[index1] == st3[index3])
{
return judge(index1 + 1, index2, index3 + 1) || judge(index1, index2 + 1, index3 + 1);
}
return false;
}
};