题目

方法

二维DP问题，二维矩阵f的大小为：s1.length() * s2.length();  f[i][j] 的状态只和f[i - 1][j] 和 f[i][j - 1]相关。

    public boolean isInterleave(String s1, String s2, String s3) {
    	int len1 = s1.length();
    	int len2 = s2.length();
    	int len3 = s3.length();
    	if (len1 + len2 != len3) {
    		return false;
    	}
    	boolean[][] f = new boolean[len1 + 1][len2 + 1];
    	f[0][0] = true;
    	for (int i = 1; i <= len1; i++) {
    		f[i][0] = f[i - 1][0] && s3.charAt(i - 1) == s1.charAt(i - 1);
    	}
    	for (int j = 1; j <= len2; j++) {
    		f[0][j] = f[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
    	}
    	
    	for (int i = 1; i <= len1; i++) {
    		for (int j = 1; j <= len2; j++) {
    			int len = i + j;
    			char c1 = s1.charAt(i - 1);
    			char c2 = s2.charAt(j - 1);
    			char c3 = s3.charAt(len - 1);
    			f[i][j] = (c1 == c3 && f[i - 1][j]) || (c2 == c3 && f[i][j - 1]);
    		}
    	}
    	return f[len1][len2];
    }