leetcode：97. 交错字符串（java 回溯）

r囧r小猫 2021-09-28 09:56 242阅读 0赞

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;
            }
        }