leetcode 097 —— Interleaving String

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：DP      

dp[i][j] 表示的是 s1的前i个字符和 s2的前j个字符 能否表示s3的前i+j个字符。
如果dp[i-1][j]==true,则只要s1的第i个字符等于s3的第i+j个字符，即s1[i-1]==s3[i+j-1],()那么就能达到前i+j个字符的匹配。

class Solution {public:bool isInterleave(string s1, string s2, string s3) {int len1 = s1.size();int len2 = s2.size();int len3 = s3.size();if (len1 + len2 != len3) return false;vector<vector<bool>> dp(len1 + 1, vector<bool>(len2 + 1, false));dp[0][0] = true;for (int i = 1; i <= len1; i++){if (dp[i - 1][0] && s1[i - 1] == s3[i - 1])dp[i][0] = true;}for (int i = 1; i <= len2; i++){if (dp[0][i - 1] && s2[i - 1] == s3[i - 1])dp[0][i] = true;}for (int i=1; i <= len1; i++){for (int j = 1; j <= len2; j++){if (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1])dp[i][j] = true;if (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1])dp[i][j] = true;}}return dp[len1][len2];}};