lintcode-medium-Interleaving String

Posted 2020-06-27 哥布林工程师

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

 

For s1 = "aabcc", s2 = "dbbca"

• When s3 = "aadbbcbcac", return true.
• When s3 = "aadbbbaccc", return false.

动态规划，用一个二维数据记录：s1前i个字符，s2前j个字符，是不是s3前i + j个字符的interleave

状态转移：

dp[i - 1][j]为true，且s1的下一个字符和s3的下一个字符相等，或者dp[i][j - 1]为true，且s2的下一个字符和s3的下一个字符相等，则dp[i][j]为true，否则dp[i][j]为false

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
        if(s1 == null && s2 == null && s3 == null)
            return true;
        if((s1 == null || s1.length() == 0) && s2.equals(s3))
            return true;
        if((s2 == null || s2.length() == 0) && s1.equals(s3))
            return true;
            
        int m = s1.length();
        int n = s2.length();
        int size3 = s3.length();
        
        if(m + n != size3)
            return false;
        
        boolean[][] dp = new boolean[m + 1][n + 1];
        
        dp[0][0] = true;
        
        for(int i = 1; i <= n; i++)
            if(s2.charAt(i - 1) == s3.charAt(i - 1) && dp[0][i - 1])
                dp[0][i] = true;
            else
                dp[0][i] = false;
        
        for(int i = 1; i <= m; i++)
            if(s1.charAt(i - 1) == s3.charAt(i - 1) && dp[i - 1][0])
                dp[i][0] = true;
            else
                dp[i][0] = false;
        
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1))
                    dp[i][j] = true;
                else if(dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1))
                    dp[i][j] = true;
                else
                    dp[i][j] = false;
            }
        }
        
        return dp[m][n];
    }
}