97 lines
3.1 KiB
Java
97 lines
3.1 KiB
Java
//给定一个经过编码的字符串,返回它解码后的字符串。
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//
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// 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。
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//
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// 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。
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//
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// 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。
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//
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//
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//
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// 示例 1:
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//
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// 输入:s = "3[a]2[bc]"
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//输出:"aaabcbc"
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//
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//
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// 示例 2:
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//
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// 输入:s = "3[a2[c]]"
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//输出:"accaccacc"
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//
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//
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// 示例 3:
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//
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// 输入:s = "2[abc]3[cd]ef"
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//输出:"abcabccdcdcdef"
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//
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//
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// 示例 4:
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//
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// 输入:s = "abc3[cd]xyz"
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//输出:"abccdcdcdxyz"
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//
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// Related Topics 栈 深度优先搜索
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// 👍 741 👎 0
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package leetcode.editor.cn;
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import java.util.Stack;
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//394:字符串解码
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public class DecodeString {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new DecodeString().new Solution();
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//accaccacc
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System.out.println(solution.decodeString("3[a2[c]]"));
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//aaabcbc
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System.out.println(solution.decodeString("3[a]2[bc]"));
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//abcabccdcdcdef
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System.out.println(solution.decodeString("2[abc]3[cd]ef"));
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//abccdcdcdxyz
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System.out.println(solution.decodeString("abc3[cd]xyz"));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public String decodeString(String s) {
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Stack<String> stack = new Stack<>();
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StringBuilder str = new StringBuilder();
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if (s.length() == 0) {
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return "";
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}
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for (int i = 0; i < s.length(); i++) {
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if (stack.isEmpty() && s.charAt(i) >= 'a' && s.charAt(i) <= 'z') {
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str.append(s.charAt(i));
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} else if (s.charAt(i) == ']') {
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StringBuilder temp = new StringBuilder();
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while (!"[".equals(stack.peek())) {
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temp.insert(0, stack.pop());
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}
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stack.pop();
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String numString = "";
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while (!stack.isEmpty() && Character.isDigit(stack.peek().charAt(0))) {
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numString = stack.pop() + numString;
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}
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int num = Integer.parseInt(numString);
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String shortStr = "";
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for (int j = 0; j < num; j++) {
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shortStr = temp + shortStr;
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}
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if (stack.isEmpty()) {
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str.append(shortStr);
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} else {
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stack.push(shortStr);
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}
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} else {
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stack.push(s.substring(i, i + 1));
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}
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}
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return str.toString();
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |