Give My Text Back

Description:
题目1 : Give My Text Back

To prepare for the English exam Little Ho collected many digital reading materials. Unfortunately the materials are messed up by a malware.

It is known that the original text contains only English letters (a-zA-Z), spaces, commas, periods and newlines, conforming to the following format:

  1. Each sentence contains at least one word, begins with a letter and ends with a period.
  2. In a sentence the only capitalized letter is the first letter.
  3. In a sentence the words are separated by a single space or a comma and a space.
  4. The sentences are separated by a single space or a single newline.

It is also known the malware changes the text in the following ways:

  1. Changing the cases of letters.
  2. Adding spaces between words and punctuations.

Given the messed text, can you help Little Ho restore the originaltext?

Input:

A string containing no more than 8192 English letters (a-zA-Z), spaces, commas, periods and newlines which is the messed text.

Output:

The original text.

Sample input:

my Name is Little Hi.
His name IS Little ho , We are friends.

Sample output:

My name is little hi.
His name is little ho, we are friends.

水题。

分析：

首先弄清楚，text只会发生两种变化：

  1. 大小写变化。
  2. word与word之间，或者word与punctuation之间增加space，space只会增加不会减少。所以不会出现 hias.HIS 这样的情况（本来应该是这样 hias. His ）。

思路：

  1. 首先，按行读取数据（这里存成string类型）。可以每读取一行，紧接着对其进行处理；也可以读取所有数据后（存储到 vector<string> 中），遍历每一项并对其处理，这里的代码采用后者。
  2. 对每一行的数据用空格' ' 进行split，得到 vector<string> 存储split后的子串，
  3. 遍历上面得到的vector<string> 对每一个子串进行分情况处理。

注意：

  1. 代码行处，对数据分割后得到的字串可能情况有下面几种情况。这里需要特殊对待的是最后一种情况。
      如果得到的子串是 hias. HIS ，处理后正确的结果应该是hias. His ，但是如果直接分割、小写化后结果是hias. his。原因是，这里判断首字母是否大写是根据 . 判断的，如果遇到了. 就设置下面子串的首字母大写，所以这里需要判断子串中是否存在 . 。
      倒数第二种情况可以直接将其转换为小写形式后直接append到结果string中，并不会产生影响。
    • ""
    • ","
    • "."
    • "hELLo"
    • "hELLo,"
    • "hELLo."
  2. 另外，上面是一开始分析漏掉的情况，提交一直wa，wa的那叫一个心痛，所以做题一定分析完全之后下手。
  3. 这道题目页可以逐个字符的读取，并对其进行处理，这种情况下会相对比本post中提到的情况容易分析和实现。参考代码

代码如下：

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#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <locale>
#include <algorithm>

using namespace std;

class Solution {
private:
    // self-definition split function
    vector<string> &split(const string &s, char delim, vector<string> &elems) {
        stringstream ss(s);
        string item;
        while (getline(ss, item, delim)) {
            elems.push_back(item);
        }
        return elems;
    }

    vector<string> split(const string &s, char delim) {
        vector<string> elems;
        split(s, delim, elems);
        return elems;
    }

public:
    string process(const string s) {
        string originSen;
        bool isCapital = false;
        vector<string> rs;
        char delim = ' ';

        rs = split(s, delim); //split the data using delim.

        for (int i = 0; i < rs.size(); i++) {
            if (rs[i] != "") {
                if (rs[i] == ","){
                    originSen.append(",");
                }
                else if (rs[i] == "."){
                    originSen.append(".");
                    isCapital = true;
                }
                else {
                    transform(rs[i].begin(), rs[i].end(), rs[i].begin(), ::tolower);
                    if (isCapital || i == 0) {
                        locale loc;
                        rs[i][0] = toupper(rs[i][0], loc);
                        if (i != 0) originSen.append(" ");
                        originSen.append(rs[i]);
                    }
                    else {
                        originSen.append(" ");
                        originSen.append(rs[i]);
                    }

                    // what mentioned in Attenton 1(注意1)
                    if (rs[i].find('.') != string::npos)
                        isCapital = true;
                    else
                        isCapital = false;
                }
            }
        }

        return originSen;
    }
};

int main() {
    vector<string> data;
    string tem;
    Solution s;

    getline(cin, tem);
    while (tem != "") {
        data.push_back(tem);
        getline(cin, tem);
    }

    for (int i = 0; i < data.size(); i++) {
        string rs = s.process(data[i]);
        cout << rs << endl;
    }

    //system("pause");
    return 0;
}