leetcode 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.

题目来源：https://leetcode.com/problems/interleaving-string/

分析

动态规划。构造二维辅助数组dp[len1+1][len2+1]。

dp[i+1][j+1] == true 表示s1[0…i]和s2[0…j]能够构造成s3[0…i+j+1]。

递推公式：

dp[i+1][j+1] = (dp[i+1][j] && (s3.at(i+j+1) == s2.at(j))) || (dp[i][j+1] && s3.at(i+j+1) == s1.at(i));

代码

[code]class Solution {
public:

    bool isInterleave(string s1, string s2, string s3) {
        int len1 = s1.length();
        int len2 = s2.length();
        int len3 = s3.length();
        if(len3 != len1 + len2)
            return false;

        vector<vector<bool> > dp(len1 + 1, vector<bool> (len2 + 1, false));
        dp[0][0] = true;

        for(int i = 0; i < len1; i++)
            dp[i+1][0] = (s1.at(i) == s3.at(i)) && dp[i][0];

        for(int i = 0; i < len2; i++)
            dp[0][i+1] = (s2.at(i) == s3.at(i)) && dp[0][i];

        for(int i = 0; i < len1; i++){
            for(int j = 0; j < len2; j++){
                dp[i+1][j+1] = (dp[i+1][j] && (s3.at(i+j+1) == s2.at(j))) || (dp[i][j+1] && s3.at(i+j+1) == s1.at(i));
            }
        }

        return dp[len1][len2];
    }
};