LeetCode75——Day24

文章目录

一、题目

2390. Removing Stars From a String

You are given a string s, which contains stars *.

In one operation, you can:

Choose a star in s.

Remove the closest non-star character to its left, as well as remove the star itself.

Return the string after all stars have been removed.

Note:

The input will be generated such that the operation is always possible.

It can be shown that the resulting string will always be unique.

Example 1:

Input: s = "leet**cod*e"

Output: "lecoe"

Explanation: Performing the removals from left to right:

  • The closest character to the 1st star is 't' in "leet**code". s becomes "leecod*e".
  • The closest character to the 2nd star is 'e' in "leecode". s becomes "lecod*e".
  • The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe".
    There are no more stars, so we return "lecoe".
    Example 2:

Input: s = "erase*****"

Output: ""

Explanation: The entire string is removed, so we return an empty string.

Constraints:

1 <= s.length <= 105

s consists of lowercase English letters and stars *.

The operation above can be performed on s.

二、题解

利用栈解决问题

cpp 复制代码
class Solution {
public:
    string removeStars(string s) {
        int n = s.length();
        stack<char> st;
        string res = "";
        for(int i = 0;i < n;i++){
            char c = s[i];
            if(c != '*') st.push(c);
            else if(c == '*' && !st.empty()) st.pop();
        }
        while(st.size()) res += st.top(),st.pop();
        reverse(res.begin(),res.end());
        return res;
    }
};
相关推荐
码客-ygluu3 小时前
一个通用游戏数据服务器
c++·架构
another heaven3 小时前
【C/C++ Ranges 是零开销抽象吗?拆开管道、惰性求值和编译期膨胀】
c语言·开发语言·c++
Code_Solitude3 小时前
C语言动态内存 + 柔性数组 + 数组名退化
c语言·算法·柔性数组
上海潭龙东海科技有限公司3 小时前
客运站管控之动态孪生技术方案(新增一屏管控模块
数据仓库·人工智能·算法·重构·动态规划
文丑颜不良啊4 小时前
LeetCode2821~2840题目解析
java·leetcode
YLCHUP6 小时前
【思维&图论建模】题解:债权(debt)_图论_建模_思路转化_思维题目_C++算法竞赛
c++·笔记·算法·数学建模·贪心算法·深度优先·图论
xiaoye-duck6 小时前
《C++ 项目:手写大模型接入 SDK》会话模块整合:内存 SessionManager 与 DataManager 持久化联动
开发语言·数据库·c++·语言模型
思麟呀6 小时前
背包DP(01背包+完全背包)
c++·算法
笑容掩饰我的失落7 小时前
区块链从小白到精通(七):Solidity 语法精讲——数据类型、函数、修饰器与事件
算法·区块链·php·智能合约·哈希算法