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_394_DecodeString.java
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package leetcode_1To300;
import java.util.Stack;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _394_DecodeString {
/**
* 394. Decode String
* s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef"
time : O(n);
space : O(n);
* @param s
* @return
*/
public String decodeString(String s) {
if (s == null || s.length() == 0) return s;
String res = "";
Stack<Integer> countStack = new Stack<>();
Stack<String> resStack = new Stack<>();
int idx = 0;
while (idx < s.length()) {
if (Character.isDigit(s.charAt(idx))) {
int count = 0;
while (Character.isDigit(s.charAt(idx))) {
count = count * 10 + (s.charAt(idx) - '0');
idx++;
}
countStack.push(count);
} else if (s.charAt(idx) == '[') {
resStack.push(res);
res = "";
idx++;
} else if (s.charAt(idx) == ']') {
StringBuilder temp = new StringBuilder(resStack.pop());
int time = countStack.pop();
for (int i = 0; i < time; i++) {
temp.append(res);
}
res = temp.toString();
idx++;
} else {
res += s.charAt(idx++);
}
}
return res;
}
}