394. Decode String

Question description

Screenshot 2016-10-05 20.47.58.png

My code

public class Solution {
    public String decodeString(String s) {
        StringBuilder sb = new StringBuilder(s);
        for (int i = 0; i < sb.length(); i++) {
            char c1 = sb.charAt(i);
            if (c1 == '[') {
                for (int j = i + 1; j < sb.length(); j++) {
                    char c2 = sb.charAt(j);
                    if (c2 == ']') {
                        int begin = 0;
                        for (int k = i - 1; k >= 0; k --) {
                            if (sb.charAt(k) > 57) {
                                begin = k + 1;
                                break;
                            }
                        }
                        int num = Integer.parseInt(sb.substring(begin, i));
                        String sub = sb.substring(i + 1, j);
                        StringBuilder tmp = new StringBuilder();
                        for (int k = 0; k < num; k++) {
                            tmp.append(sub);
                        }
                        sb.replace(begin, j + 1, tmp.toString());
                        i = 0;
                        break;
                    } else if (c2 == '[') {
                        break;
                    }
                }
            }
        }
        return sb.toString();
    }
}

Solution

For brackets that have no inside brackets, directly multiply the content in the bracket by the number before the bracket. Otherwise, look into the bracket and get the bracket without inner brackets.

Test result

Screenshot 2016-10-05 20.47.38.png
最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

推荐阅读更多精彩内容