HJ26 String sorting ●●

HJ26 String sorting ●●

describe

Write a program to sort the characters in the input string according to the following rules.

Rule 1: English letters are arranged from A to Z, case insensitive.

For example, input: Type output: epTy

Rule 2: When the upper and lower case of the same English letter exists at the same time, it will be arranged according to the input order.

For example, input: BabA output: aABb

Rule 3: Other characters that are not English letters remain in their original positions.

For example, input: By?e output: Be?y

Data range: the length of the input string satisfies 1 \le n \le 1000 \1≤n≤1000

Enter a description:

input string

Output description:

output string

example

Song:A Famous Saying: Much Ado About Nothing (2012/8).Song
:A aaAAbc dFgghh: iimM nNn oooos Sttuuuy (2012/8).

answer

1. Bucket sort

Count the order of appearance of 26 letters, and perform traversal filling.
Time complexity: O(n)
Space complexity: O(1)

#include <iostream>
#include <string>
#include <vector>
using namespace std;

bool isLetter(char ch){
    
    
    return toupper(ch) <= 'Z' && toupper(ch) >= 'A';
}

void bucketSort(string& str){
    
    
    vector<string> buckets(26, "");
    for(char ch : str){
    
    
        if(isLetter(ch)){
    
    
            buckets[toupper(ch)-'A'].push_back(ch);		// 对26个字母进行记录，有相对顺序
        }
    }
    int str_idx = 0, buck_idx = 0, num_idx = 0;
    while(str_idx < str.length()){
    
    
        if(isLetter(str[str_idx])){
    
    
            if(buck_idx < buckets[num_idx].size()){
    
    		// 填充当前字母
                str[str_idx++] = buckets[num_idx][buck_idx++];	
            }else{
    
    				// 当前字母遍历填充完成，跳到下一个字母
                buck_idx = 0;
                ++num_idx;		
            }
        }else{
    
    
            ++str_idx;			// 非字母位置，跳过
        }
    }    
}

int main(){
    
    
    string str;
    getline(cin, str);
    bucketSort(str);
    cout << str;
    return 0;
}

2. Merge sort

The idea of ​​merge sorting is to judge and skip non-letter positions,
nlogn complexity requires additional space, and the recursion times out when the data is large.

#include <iostream>
#include <string>
#include <vector>
using namespace std;

bool isLetter(char ch){
    
    
    return toupper(ch) <= 'Z' && toupper(ch) >= 'A';
}

void mergeSort(string& str, string& sorted, int start, int end){
    
    
    if(start >= end) return;
    int mid = start + (end - start) / 2;
    mergeSort(str, sorted, start, mid);
    mergeSort(str, sorted, mid+1, end);
    int left = start, right = mid+1;
    int idx = start;
    while(idx <= end){
    
    
        if(isLetter(str[left]) == false){
    
    
            ++left;
        }else if(isLetter(str[right]) == false){
    
    
            ++right;
        }else if(isLetter(str[idx]) == false){
    
    
            ++idx;
        }else if(left > mid){
    
    
            sorted[idx++] = str[right++];
        }else if(right > end){
    
    
            sorted[idx++] = str[left++];
        }else if(toupper(str[left]) > toupper(str[right])){
    
    
            sorted[idx++] = str[right++];
        }else if(toupper(str[left]) < toupper(str[right])){
    
    
            sorted[idx++] = str[left++];
        }else{
    
    
            sorted[idx++] = str[left++];
        }
    }
    for(int i = start; i <= end; ++i) str[i] = sorted[i];
}

int main(){
    
    
    string str;
    getline(cin, str);
    string sorted = str;
    mergeSort(str, sorted, 0, str.length()-1);
    cout << str;
    return 0;
}