leetcode：97. 交错字符串（java 回溯 ）

给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。
示例 1:
输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true
示例 2:
输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false

第1种方法回溯法时间复杂度肯定是很不理想的

  public boolean isInterleave(String s1, String s2, String s3) {
        if(s3.length()!=s1.length()+s2.length())return false;
        return dfs(0, 0, 0, s1, s2, s3);
    }

    public boolean dfs(int index, int s1i, int s2i, String s1, String s2, String s3) {
        if (index == s3.length() && s1i == s1.length() && s2i == s2.length()) {
            return true;
        }
        if (s1i > s1.length() || s2i > s2.length() || index > s3.length()) {
            return false;
        }
        char s1c = '!';
        char s2c = '!';
        char s3c = '!';
        if (s1i < s1.length()) {
            s1c = s1.charAt(s1i);
        }
        if (s2i < s2.length()) {
            s2c = s2.charAt(s2i);
        }
        if (index < s3.length()) {
            s3c = s3.charAt(index);
        }
        //两个字符都相等 分两种情况进行回溯
        if (s1c == s3c && s2c == s3c) {
            if (dfs(index + 1, s1i + 1, s2i, s1, s2, s3)) {
                return true;
            } else {
                return dfs(index + 1, s1i, s2i + 1, s1, s2, s3);
            }
        } else if (s1c == s3c) {
            return dfs(index + 1, s1i + 1, s2i, s1, s2, s3);
        } else if (s2c == s3c) {
            return dfs(index + 1, s1i, s2i + 1, s1, s2, s3);
        } else {
            return false;
        }
    }