Leetcode: Mini Parser
Given a nested list of integers represented as a string, implement a parser to deserialize it. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Note: You may assume that the string is well-formed: String is non-empty.
String does not contain white spaces.
String contains only digits 0-9, [, - ,, ].
Example 1: Given s = "324", You should return a NestedInteger object which contains a single integer 324.
Example 2: Given s = "[123,[456,[789]]]", Return a NestedInteger object containing a nested list with 2 elements: 1. An integer containing value 123.
2. A nested list containing two elements:
i. An integer containing value 456.
ii. A nested list with one element:
a. An integer containing value 789.
有括号这种一般要用stack, stack top 就是当前着眼的那一个NestedInteger, 可以对其添加新的元素。
注意47行判断很关键,顺带处理了 "[]," 括号后面是逗号的情况
/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* public interface NestedInteger {
* // Constructor initializes an empty nested list.
* public NestedInteger();
*
* // Constructor initializes a single integer.
* public NestedInteger(int value);
*
* // @return true if this NestedInteger holds a single integer, rather than a nested list.
* public boolean isInteger();
*
* // @return the single integer that this NestedInteger holds, if it holds a single integer
* // Return null if this NestedInteger holds a nested list
* public Integer getInteger();
*
* // Set this NestedInteger to hold a single integer.
* public void setInteger(int value);
*
* // Set this NestedInteger to hold a nested list and adds a nested integer to it.
* public void add(NestedInteger ni);
*
* // @return the nested list that this NestedInteger holds, if it holds a nested list
* // Return null if this NestedInteger holds a single integer
* public List<NestedInteger> getList();
* }
*/
public class Solution {
public NestedInteger deserialize(String s) {
if (!s.startsWith("[")) {
return new NestedInteger(Integer.parseInt(s));
}
NestedInteger res = new NestedInteger();
Stack<NestedInteger> st = new Stack<NestedInteger>();
st.push(res);
int start = 1;
for (int i=1; i<s.length(); i++) {
char c = s.charAt(i);
if (c == '[') {
NestedInteger ni = new NestedInteger();
st.peek().add(ni);
st.push(ni);
start = i+1;
}
else if (c==',' || c==']') {
if (i > start) { // "],"这种情况
int val = Integer.parseInt(s.substring(start, i));
st.peek().add(new NestedInteger(val));
}
start = i+1;
if (c == ']') st.pop();
}
}
return res;
}
}
Leetcode: Mini Parser的更多相关文章
- [LeetCode] Mini Parser 迷你解析器
Given a nested list of integers represented as a string, implement a parser to deserialize it. Each ...
- 【LeetCode】385. Mini Parser 解题报告(Python)
[LeetCode]385. Mini Parser 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/mini-parser/ ...
- 385. Mini Parser - LeetCode
Question 385. Mini Parser Solution 分析:用NI(count,list)来表示NestedInteger,则解析字符串[123,[456,[789]]]过程如下: # ...
- 385 Mini Parser 迷你解析器
Given a nested list of integers represented as a string, implement a parser to deserialize it.Each e ...
- [Swift]LeetCode385. 迷你语法分析器 | Mini Parser
Given a nested list of integers represented as a string, implement a parser to deserialize it. Each ...
- 385. Mini Parser
括号题一般都是stack.. 一开始想的是存入STACK的是SRING,然后POP出括号在构建新的NestedInteger放到另一个里面,但是操作起来费时费力. 后来猛然发现其实可以直接吧Neste ...
- LeetCode All in One 题目讲解汇总(持续更新中...)
终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance ...
- leetcode bugfree note
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的 ...
- leetcode & lintcode for bug-free
刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be it ...
随机推荐
- JS初学者必备的几个经典案例(二)!!!
一.写出当前年份的前后5年的日期表 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" " ...
- Blender to XPS(blender 2.7x Internal materials)
Things we are gonna need are Blender 2.7x www.blender.org/ XPS tools addon for Blender A model made ...
- nrf51822裸机教程-GPIO
首先看看一下相关的寄存器说明 Out寄存器 输出设置寄存器 每个比特按顺序对应每个引脚,bit0对应的就是 引脚0 该寄存器用来设置 引脚作为输出的时候的 输出电平为高还是低. 与输出设置相关的 还有 ...
- iftop ifstat
ifstat 介绍 ifstat工具是个网络接口监测工具,比较简单看网络流量 实例 默认使用 #ifstat eth0 eth1 KB/s in KB/s out KB/s in KB/s out 0 ...
- Qt Focus事件,FocusInEvent()与FocusOutEvent()
描述:一开始我要实现的目的就是,在一个窗体上有多个可编辑控件(比如QLineEdit.QTextEdit等),当哪个控件获得焦点,哪个控件的背景就高亮用来起提示作用,查了下文档应该用focusInEv ...
- C语言解析json类型数据
转自:http://buluzhai.iteye.com/blog/845404 首先感谢作者!! 我使用的是cJSON:http://sourceforge.net/projects/cjson ...
- [LeetCode] Combinations (bfs bad、dfs 递归 accept)
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. For exampl ...
- [LeetCode]题解(python):098 Validate Binary Search Tree
题目来源 https://leetcode.com/problems/validate-binary-search-tree/ Given a binary tree, determine if it ...
- 计算A+B及其结果的标准形式输出
题目: 代码链接 解题思路: 首先,读懂题目,题目要求我们计算两个整型数a,b之和,这是简单的加法计算,与平常的题目一般无二.但是此题的不同在于要求我们输出的数必须是标准形式,题目也对标准形式做了相应 ...
- imx6 android5.1 打开 调试串口
imx6的工板烧录android 5.1的镜像,uboot中能使用debug口,kernel,文件系统中不能使用debug口. 打开kenel和文件系统debug口方法,在uboot的bootargs ...