题解 | #字符串排序#

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String originalStr = scan.nextLine();
        StringBuffer sb = new StringBuffer("");
        int sign = 0;
        int start = 0;
        char[] chrs = originalStr.toCharArray();
        Queue<Character> queue = new LinkedList<>();
        for (int i = 0; i < chrs.length; i++) {
            char chr = chrs[i];
            if ((chr >= 'A' && chr <= 'Z') || (chr >= 'a' && chr <= 'z')) {
                if (sign == 0) {
                    sign = 1;
                    start = i;
                }
            } else {
                if (sign == 1) {
                    sign = 0;
                    sb.append(originalStr.substring(start, i));
                }
                queue.add(chr);
            }
        }
        if (sign == 1) {
            sb.append(originalStr.substring(start));
        }
        char[] newchrs = process(new String(sb)).toCharArray();
        int index = 0;
        StringBuffer ans = new StringBuffer("");
        for (int i = 0; i < chrs.length; i++) {
            char chr = chrs[i];
            if ((chr >= 'A' && chr <= 'Z') || (chr >= 'a' && chr <= 'z')) {
                ans.append(newchrs[index++]);
            } else {
                ans.append(queue.poll());
            }
        }
        System.out.println(ans);
    }
    public static String process(String str) {
        char[] chrs = str.toCharArray();
        // quickSort(chrs); // 快速排序具有不稳定性，所以不能使用
        mergeSort(chrs); // 归并排序具有稳定性，因此我们选择它
        StringBuffer sb = new StringBuffer("");
        for (char chr : chrs) {
            sb.append(chr);
        }
        return new String(sb);
    }
    /**********************************************************************************/
    public static void quickSort(char[] chrs) {
        if (null == chrs || chrs.length < 2) {
            return;
        }
        sort(chrs, 0, chrs.length - 1);
    }
    public static void sort(char[] chrs, int start, int end) {
        if (start >= end) {
            return;
        }
        int l = start - 1;
        int r = end + 1;
        int p = start;
        char compareChr = String.valueOf(chrs[end]).toLowerCase().charAt(0);
        while (p < r) {
            if (String.valueOf(chrs[p]).toLowerCase().charAt(0) < compareChr) {
                swap(chrs, p++, ++l);
            } else if (String.valueOf(chrs[p]).toLowerCase().charAt(0) > compareChr) {
                swap(chrs, p, --r);
            } else {
                p++;
            }
        }
        sort(chrs, start, l);
        sort(chrs, r, end);
    }
    public static void swap(char[] chrs, int p1, int p2) {
        char tmp = chrs[p1];
        chrs[p1] = chrs[p2];
        chrs[p2] = tmp;
    }
    /**********************************************************************************/
    public static void mergeSort(char[] chrs) {
        if (null == chrs || chrs.length < 2) {
            return;
        }
        process1(chrs, 0, chrs.length - 1);
    }
    public static void process1(char[] chrs, int start, int end) {
        if (start >= end) {
            return;
        }
        int mid = start + ((end - start) >> 1);
        process1(chrs, start, mid);
        process1(chrs, mid + 1, end);
        merge(chrs, start, mid, end);
    }
    public static void merge(char[] chrs, int start, int mid, int end) {
        char[] helper = new char[end - start + 1];
        int index = 0;
        int p1 = start;
        int p2 = mid + 1;
        while (p1 <= mid && p2 <= end) {
            helper[index++] = String.valueOf(chrs[p1]).toLowerCase().charAt(0) <= String.valueOf(chrs[p2]).toLowerCase().charAt(0) ? chrs[p1++] : chrs[p2++];
        }
        while (p1 <= mid) {
            helper[index++] = chrs[p1++];
        }
        while (p2 <= end) {
            helper[index++] = chrs[p2++];
        }
        for (int i = 0; i < helper.length; i++) {
            chrs[start + i] = helper[i];
        }
    }
}