标签：
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.
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思路：
s1 = "aabcc";
s2 = "dbbca";
s3 = "aadbbcbcac";
1）如果i==0 && j==0,dp[0][0] = true；
2）如果i==0,判断s2[j]==s3[i+j] 并且 dp[i][j-1]==true，则dp[i][j] = true；
3）如果j==0,与2）类似；
4）如果i != 0 && j != 0，判断(dp[i][j-1] && s2[j]==s3[i+j]) || (dp[i-1][j] && s1[i]==s3[i+j])；
5）dp[s1.length()][s2.length()]即为结果。
过程如下图所示：
| j | 0 | 1 | 2 | 3 | 4 | 5 | |
| i | s2 | d | b | b | c | a | |
| 0 | s1 | T | F | F | F | F | F |
| 1 | a | T | F | F | F | F | F |
| 2 | a | T | T | T | T | T | F |
| 3 | b | F | T | T | F | T | F |
| 4 | c | F | F | T | T | T | T |
| 5 | c | F | F | F | T | F | T |
c++ code:
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int len1 = s1.length();
int len2 = s2.length();
bool dp[len1+1][len2+1];
for(int i=0;i<=len1;i++) {
for(int j=0;j<=len2;j++) {
if(i==0 && j==0)
dp[i][j] = true;
else if(i==0)
dp[i][j] = (dp[i][j-1] && s2[j-1]==s3[i+j-1]);
else if(j==0)
dp[i][j] = (dp[i-1][j] && s1[i-1]==s3[i+j-1]);
else
dp[i][j] = (dp[i][j-1] && s2[j-1]==s3[i+j-1])||(dp[i-1][j] && s1[i-1]==s3[i+j-1]);
}
}
return dp[len1][len2];
}
};标签：
原文地址：http://blog.csdn.net/itismelzp/article/details/51505419
