LeetCode #14 最长公共前缀--简单

编写一个函数来查找字符串数组中的最长公共前缀。
如果不存在公共前缀，返回空字符串 “”。
示例 1:

输入: ["flower","flow","flight"]
输出: "fl"

示例 2:

输入: ["dog","racecar","car"]
输出: ""
解释: 输入不存在公共前缀。

示例3:

[]

示例4:

[""]

示例5:

["a"]

示例6:

["aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "a"]

说明:
所有输入只包含小写字母 a-z 。

• 解题思路：
  先考虑特殊情况，当输入为空时，有示例3和示例4两种情况；当输入字符串长度为1时，如示例5；当字符串过长时，如示例6；

  class Solution {
  public:
      string longestCommonPrefix(vector<string>& strs) {
          string result = "";
          string str, temp;
          int length;
          if (strs.size() == 0) // 考虑 [] 与 [""] 两种情况
              return result;
          else if (strs.size() == 1){ // 考虑 ["a"] 
              return strs[0];
          }
          else{
              str = strs[0];
              for (int i=1; i<strs.size(); ++i){
                  temp = strs[i];
                  result = "";
                  if (str.length() < temp.length())
                      length = str.length();
                  else
                      length = temp.length();
                  for (int j=0; j<length; ++j){
                      if (temp[j] == str[j]){
                          result += temp[j];
                          //continue;
                      }
                      else
                          break;
                  }
                  str = result;
              }
              return result;
          }
      }
  };
  

Tips:
vector\<string\> strs 的字符串长度 strs.size()
string str字符差un长度str.length()