【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.

题意：s1和s2是否能构成s3

思路：动态规划问题

dp[i][j]表示S1前i个字符与S2前j个字符是否构成S3前i+j字符；

class Solution{public:bool isInterleve(string s1, string s2, string s3){int rows = s1.length();int cols = s2.length();int lens = s3.length();if ((rows+cols)!=lens){return false;}if (rows==0 || cols==0){if ((rows==0 && s2!=s3)|| (cols==0 && s1!=s3)){return false;}}vector<vector<int>> dp(rows+1, vector<int>(cols+1, false));dp[0][0] = true;   //初始化操作for (int i=1; i<=rows; i++){dp[i][0] = dp[i - 1][0] && (s1[i-1]==s3[i-1]);}for (int i = 1; i <= cols; i++){dp[0][i] = dp[0][i] && (s2[i - 1] == s3[i - 1]);}for (int i=1; i<=rows; i++){for (int j=1; j<=cols; j++){//要满足条件，则s3的最后一个要么是s1的最后一个字符//要么是s2的最后一个字符if (s3[i + j - 1] == s1[i - 1] && dp[i - 1][j]){dp[i][j] = true;}else if (s3[i + j - 1] == s2[j - 1] && dp[i][j - 1]){dp[i][j] = true;}elsedp[i][j] = false;}}return dp[rows][cols];}};

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.