最长公共前缀短语

题目：最长公共子串的变种。给出一系列语句，要找出最长的公共前缀短语。E.g. "test a short phrase", "test a slightly longer phrase" -> "test a"

需要注意的是前缀，所以都是从串首开始。短语，所以最长前缀短语是"test a"，而不是"test a s“。

• 对于vector<string>::const_iterator strIT1，使用length()方法时，要用(*strIT1).length()，用括号括起来。
• 注意find()函数查找成功时返回所在位置，失败返回string::npos的值。
• operator[]和at()均返回当前字符串中第n个字符的位置，但at函数提供范围检查，当越界时会抛出out_of_range异常，下标运算符[]不提供检查访问。
• operator[]和at()返回值为char，所以如果要跟空格比较的话，要用' '。
• string - C++ Reference
  • 　　http://www.cplusplus.com/reference/string/string/
• c++中string类的详解 - yzl_rex - 博客频道 - CSDN.NET
  • 　　http://blog.csdn.net/yzl_rex/article/details/7839379
• 字符串（C++ STL <字符串>）
  • 　　https://msdn.microsoft.com/zh-cn/library/y4k49tt9(v=vs.140).aspx

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

class Solution 
{
public:
        string substr(const vector<string>& vecStr)
        {
            int maxLength = 0;
            string sLongestComm = "";
        
            // Corner case
            if (vecStr.size() == 0)     return "";
            
            for (vector<string>::const_iterator strIT1 = vecStr.begin(); strIT1 != vecStr.end(); strIT1 ++)
            {
                // Stop if it is shorter than the current longest common phrase
                // Need to use (*strIT1).length() rather than *strIT1.length()
                for (int i = (*strIT1).length(); i > maxLength; i --)
                {
                    string sComm = (*strIT1).substr(0, i);

                    // Use at() rather than [] as it is exception safety                
                    if ((i != (*strIT1).length()) && (*strIT1).at(sComm.length()) != ' ')
                       continue;
                
                    bool isComm = true;
                
                    for (vector<string>::const_iterator strIT2 = vecStr.begin(); strIT2 != vecStr.end(); strIT2 ++)
                    {
                        // Remember the way to check not found result
                        if ((*strIT2).find(sComm) == string::npos)
                        {
                            isComm = false;
                            break;
                        }            
                    }
                    
                    if (isComm && (sComm.length() > maxLength) && (((*strIT1).at(sComm.length()) == ' ') || (i == (*strIT1).length())))
                    {
                        maxLength = sComm.length();
                        sLongestComm = sComm;    
                    }
                } // end of for
            } // end of for
            
            return sLongestComm;
        }
};

int main ()
{
    Solution testSolution;
    string array[] = {"test a short phrase", "test a slightly longer phrase"};
    vector<string> vecArray(array, array + 2);
    
    for (vector<string>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
        cout << *vecCIT << endl;
    cout << endl;

    cout << testSolution.substr(vecArray);
    
    getchar();
    
    return 0;
}

//
//  main.cpp
//  LeetCode
//
//  Created by Hao on 2017/3/16.
//  Copyright © 2017年 Hao. All rights reserved.
//

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

class Solution
{
public:
    string substr(const vector<string>& vecStr)
    {
        int maxLength = 0;
        string sLongestComm = "";
        
        // Corner case
        if (vecStr.size() == 0)     return "";
        
        for (auto j = 0; j < vecStr.size(); j ++)
        {
            // Stop if it is shorter than the current longest common phrase
            // Need to use (*strIT1).length() rather than *strIT1.length()
            for (auto i = vecStr[j].size(); i > maxLength; i --)
            {
                string sComm = vecStr[j].substr(0, i);
                
                // Use at() rather than [] as it is exception safety
                if ((i != vecStr[j].length()) && vecStr[j].at(sComm.length()) != ' ')
                    continue;
                
                bool isComm = true;
                
                for (auto k = 0; k < vecStr.size(); k ++)
                {
                    // Remember the way to check not found result
                    if (vecStr[k].find(sComm) == string::npos)
                    {
                        isComm = false;
                        break;
                    }
                }
                
                if (isComm && (sComm.length() > maxLength) && ((vecStr[j].at(sComm.length()) == ' ') || (i == vecStr[j].length())))
                {
                    maxLength = sComm.length();
                    sLongestComm = sComm;
                }
            } // end of for
        } // end of for
        
        return sLongestComm;
    }
};

int main ()
{
    Solution testSolution;
    string array[] = {"test a short phrase", "test a slightly longer phrase", "test a slightly slightly longer phrase", "test "};
    vector<string> vecArray(array, array + 4);
    
    for (auto vecCIT : vecArray)
        cout << vecCIT << endl;
    cout << endl;
    
    cout << testSolution.substr(vecArray) << endl;
    
    return 0;
}