leet-code/src/main/java/leetcode/editor/cn/DecodeString.java
2021-04-29 23:21:52 +08:00

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//给定一个经过编码的字符串,返回它解码后的字符串。
//
// 编码规则为: k[encoded_string],表示其中方括号内部的 encoded_string 正好重复 k 次。注意 k 保证为正整数。
//
// 你可以认为输入字符串总是有效的;输入字符串中没有额外的空格,且输入的方括号总是符合格式要求的。
//
// 此外,你可以认为原始数据不包含数字,所有的数字只表示重复的次数 k ,例如不会出现像 3a 或 2[4] 的输入。
//
//
//
// 示例 1
//
// 输入s = "3[a]2[bc]"
//输出:"aaabcbc"
//
//
// 示例 2
//
// 输入s = "3[a2[c]]"
//输出:"accaccacc"
//
//
// 示例 3
//
// 输入s = "2[abc]3[cd]ef"
//输出:"abcabccdcdcdef"
//
//
// 示例 4
//
// 输入s = "abc3[cd]xyz"
//输出:"abccdcdcdxyz"
//
// Related Topics 栈 深度优先搜索
// 👍 741 👎 0
package leetcode.editor.cn;
import java.util.Stack;
//394:字符串解码
public class DecodeString {
public static void main(String[] args) {
//测试代码
Solution solution = new DecodeString().new Solution();
//accaccacc
System.out.println(solution.decodeString("3[a2[c]]"));
//aaabcbc
System.out.println(solution.decodeString("3[a]2[bc]"));
//abcabccdcdcdef
System.out.println(solution.decodeString("2[abc]3[cd]ef"));
//abccdcdcdxyz
System.out.println(solution.decodeString("abc3[cd]xyz"));
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public String decodeString(String s) {
Stack<String> stack = new Stack<>();
StringBuilder str = new StringBuilder();
if (s.length() == 0) {
return "";
}
for (int i = 0; i < s.length(); i++) {
if (stack.isEmpty() && s.charAt(i) >= 'a' && s.charAt(i) <= 'z') {
str.append(s.charAt(i));
} else if (s.charAt(i) == ']') {
StringBuilder temp = new StringBuilder();
while (!"[".equals(stack.peek())) {
temp.insert(0, stack.pop());
}
stack.pop();
String numString = "";
while (!stack.isEmpty() && Character.isDigit(stack.peek().charAt(0))) {
numString = stack.pop() + numString;
}
int num = Integer.parseInt(numString);
String shortStr = "";
for (int j = 0; j < num; j++) {
shortStr = temp + shortStr;
}
if (stack.isEmpty()) {
str.append(shortStr);
} else {
stack.push(shortStr);
}
} else {
stack.push(s.substring(i, i + 1));
}
}
return str.toString();
}
}
//leetcode submit region end(Prohibit modification and deletion)
}