97. Interleaving String
题目
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.
解析
- LeetCode(97) Interleaving String
状态cache[i][j]表示,s1的前i个字符和s2的前j个字符是否能交叉构成s3的前i+j个字符
初始化:
cache[0][0] = True 因为两个空字符串可以组成空字符串
边界情况是一个字符串为空,初始化只需判断另一个字符串和目标字符串前x为是否相等
递推关系 cache[i][j] = (s1[i] == s3[i+j] and cache[i-1][j]) or (s2[j] == s3[i+j] and cache[i][j-1])
class Solution_97 {
public:
bool isInterleave(string s1, string s2, string s3) {
if (s3.length() != s1.length() + s2.length())
return false;
bool table[s1.length() + 1][s2.length() + 1];
for (int i = 0; i < s1.length() + 1; i++)
for (int j = 0; j < s2.length() + 1; j++){
if (i == 0 && j == 0)
table[i][j] = true;
else if (i == 0)
table[i][j] = (table[i][j - 1] && s2[j - 1] == s3[i + j - 1]);
else if (j == 0)
table[i][j] = (table[i - 1][j] && s1[i - 1] == s3[i + j - 1]);
else
table[i][j] = (table[i - 1][j] && s1[i - 1] == s3[i + j - 1]) || (table[i][j - 1] && s2[j - 1] == s3[i + j - 1]);
}
return table[s1.length()][s2.length()];
}
};
题目来源