Interleaving String 29

Question

Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1 and s2.

Example

For s1 = "aabcc", s2 = "dbbca"

When s3 = "aadbbcbcac", return true.

When s3 = "aadbbbaccc", return false.

Challenge

O(n2) time or better

Solution

dp[i][j]表示s1前i个和s2前j个对s3前i+j个是否interleaving string。

1. 首先初始化。遍历s1，初始化所有的dp[i][0]

2. 再遍历s2，初始化所有的dp[0][j]

3. 若s3的第i＋j－1位和s1的第i位相等，则看dp[i-1][j]是否为true；同理，若s3的i+j-1位和s2的第j位相等，则看dp[i][j-1]是否为true。只要两种情况中的任意一种为true，则dp[i][j]为true。

代码如下：

public class Solution {
    /**
     * Determine whether s3 is formed by interleaving of s1 and s2.
     * @param s1, s2, s3: As description.
     * @return: true or false.
     */
    public boolean isInterleave(String s1, String s2, String s3) {
        // write your code here
        int n = s1.length();
        int m = s2.length();
        boolean[][] f = new boolean[n + 1][m + 1];

        if(n + m != s3.length()){
            return false;
        }
        f[0][0] = true;

        for(int i = 1; i <= n; i++){
            if(s3.charAt(i - 1) == s1.charAt(i - 1) && f[i - 1][0]){
                f[i][0] = true;
            }
        }

        for(int j = 1; j <= m; j++){
            if(s3.charAt(j - 1) == s2.charAt(j - 1) && f[0][j - 1]){
                f[0][j] = true;
            }
        }

        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(f[i - 1][j] && (s1.charAt(i - 1) == s3.charAt(i + j - 1)) || f[i][j - 1] && (s2.charAt(j - 1) == s3.charAt(i + j - 1))){
                    f[i][j] = true;
                }
            }
        }

        return f[n][m];
    }
}