题解 | #字符串合并处理#

字符串合并处理

http://www.nowcoder.com/practice/d3d8e23870584782b3dd48f26cb39c8f

题目本身其实并不是很难,主要是代码量太多了。细心读题,一步一步实现就好。

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String[] strs = scan.nextLine().split(" ");
        char[] original = (strs[0] + strs[1]).toCharArray();
        char[] even = new char[(original.length + 1) / 2];
        char[] odd = new char[original.length / 2];
        int p1 = 0;
        int p2 = 0;
        for (int i = 0; i < original.length; i++) {
            if (i % 2 == 0) {
                even[p1++] = original[i];
            } else {
                odd[p2++] = original[i];
            }
        }
        mergeSort(even);
        mergeSort(odd);
        p1 = 0;
        p2 = 0;
        StringBuffer sb = new StringBuffer("");
        while (p1 < even.length && p2 < odd.length) {
            sb.append(even[p1]);
            sb.append(odd[p2]);
            p1++;
            p2++;
        }
        if (p1 < even.length) {
            sb.append(even[p1]);
        }
        System.out.println(transform(new String(sb)));
    }
    public static void mergeSort(char[] chrs) {
        if (null == chrs || chrs.length < 2) {
            return;
        }
        process(chrs, 0, chrs.length - 1);
    }
    public static void process(char[] chrs, int start, int end) {
        if (start >= end) {
            return;
        }
        int mid = start + ((end - start) >> 1);
        process(chrs, start, mid);
        process(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++] = chrs[p1] <= chrs[p2] ? 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];
        }
    }
    public static String transform(String str) {
        HashMap<Character, Integer> CharacterToInteger = new HashMap<>();
        HashMap<Integer, Character> IntegerToCharacter = new HashMap<>();
        for (int i = 0; i < 10; i++) {
            CharacterToInteger.put((char) ('0' + i), i);
            IntegerToCharacter.put(i, (char) ('0' + i));
        }
        for (int i = 0; i < 6; i++) {
            CharacterToInteger.put((char) ('A' + i), i + 10);
            CharacterToInteger.put((char) ('a' + i), i + 10);
            IntegerToCharacter.put(i + 10, (char) ('A' + i));
        }
        StringBuffer sb = new StringBuffer("");
        char[] chrs = str.toCharArray();
        for (char chr : chrs) {
            if ((chr >= '0' && chr <= '9') || (chr >= 'A' && chr <= 'F') || (chr >= 'a' && chr <= 'f')) {
                int val = CharacterToInteger.get(chr);
                String tmpStr = DecimalToBinaryAndReverse(val);
                int tmpVal = BinaryToDecimal(tmpStr);
                sb.append(IntegerToCharacter.get(tmpVal));
            } else {
                sb.append(chr);
            }
        }
        return new String(sb);
    }
    public static String DecimalToBinaryAndReverse(int num) {
        StringBuffer sb = new StringBuffer("");
        while (num != 0) {
            sb.append(num % 2);
            num /= 2;
        }
        while (sb.length() != 4) {
            sb.append("0");
        }
        return new String(sb);
    }
    public static int BinaryToDecimal(String str) {
        int ans = 0;
        int acc = 0;
        char[] chrs = str.toCharArray();
        for (int i = chrs.length - 1; i > -1; i--) {
            if (chrs[i] == '1') {
                ans += (int) Math.pow(2, acc);
            }
            acc++;
        }
        return ans;
    }
}
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发布于 2022-04-27 11:43

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