Minimal string 栈贪心

最新推荐文章于 2021-07-26 16:50:33 发布

转载最新推荐文章于 2021-07-26 16:50:33 发布 · 158 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/7274106.html

本文介绍了一个字符串处理问题，通过特定算法实现字符串s的重新排列，目标是得到字典序最小的字符串u。文章提供了完整的C++代码示例，详细解释了如何通过栈结构和迭代方式，实现字符串的最优重组。

Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:

Extract the first character of s and append t with this character.
Extract the last character of t and append u with this character.

Petya wants to get strings s and t empty and string u lexicographically minimal.

You should write a program that will help Petya win the game.

Input

First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters.

Output

Print resulting string u.

Examples

input

cab

output

abc

input

acdb

output

abdc

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 100009
#define N 40
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-9;
const double PI = acos(-1.0);
/*
从前往后遍历
要让字典序的顺序尽量小，那么如果该字符后面有字典序比它小的，应当一直压栈直到栈顶元素优先级更高为止。（保证字典序小的优先！）

*/
char s[MAXN],best[MAXN];
stack<char> S;
int main()
{
    scanf("%s", s);    
    int l = strlen(s);
    best[l] = 'z';
    int p = 0;
    for (int i = l - 1; i >= 0; i--)
        best[i] = min(s[i], best[i + 1]);
    while (!S.empty() || p < l)
    {
        if (!S.empty() && S.top() <= best[p])
            putchar(S.top()), S.pop();
        else
            S.push(s[p++]);
    }
    printf("\n");
    return 0;
}

转载于:https://www.cnblogs.com/joeylee97/p/7274106.html

Minimal string 栈 贪心

Minimal string 栈贪心