Interleaving String
Total Accepted: 15409 Total Submissions: 79580My Submissions
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.
题意：判断某一个字符串是否是由另外两个字符串交织构成的
思路：dp
设dp[i][j]表示s3[0...i+j)能否由s1[0...i)和s2[0...j)交织构成
dp[i][j] = dp[i-1][j] ,if s3[i + j - 1] == s1[i - 1]
dp[i][j] = dp[i][j-1] ,if s3[i + j - 1] == s2[j - 1]
dp[i][j] = false, otherwise
初始化 dp[i][j]
dp[0][k] = true , if s2[0...k) == s3[0...k), 0 <= k <= j
dp[0][k] = false, otherwise
dp[k][0] = true , if s1[0...k) == s3[0...k), 0 <= k <= i
dp[k][0] = false, otherwise
复杂度：时间O(n^2)，空间O(n^2)
bool isInterleave(string s1, string s2, string s3){
int len1 = s1.size(), len2 = s2.size(), len3 = s3.size();
if(len1 + len2 != len3) return false;
vector
> dp(len1 + 1, vector
(len2 + 1, false)); //初始化 dp[0][0] = true; for(int i = 1; i <= len1; ++i){ dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]); } for(int j = 1; j <= len2; ++j){ dp[0][j] = dp[0][j - 1] && (s2[j - 1] == s3[j - 1]); } //迭代 for(int i = 1; i <= len1; ++i){ for(int j = 1; j <= len2; ++j){ dp[i][j] = (s3[i + j - 1] == s1[i - 1] && dp[i - 1][j] || s3[i + j - 1] == s2[j - 1] && dp[i][j - 1]); } } return dp[len1][len2]; }