[LeetCode] Longest Common Prefix 解题报告

Write a function to find the longest common prefix string amongst an array of strings.» Solve this problem

[解题报告]
又一个实现题。遍历字符串数组，每次用当前prefix和下一个字符串匹配以生成新的prefix。

[code]1:    string longestCommonPrefix(vector<string> &strs) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      string compare;
5:      if(strs.size() == 0) return compare;
6:      compare = strs[0];
7:      for(int i =1; i< strs.size() ; i++)
8:      {
9:        string prefix;
10:        int k =0;
11:        while(k< compare.size() && k< strs[i].size())
12:        {
13:          if(compare[k] != strs[i][k])
14:            break;
15:          prefix.append(1, compare[k]);
16:          k++;
17:        }
18:        compare.clear();
19:        compare.append(prefix.c_str());
20:      }
21:      return compare;
22:    }

另一个直观的解法就是对于每一个字母比较所有字符串，直到遇到任何一个不匹配。这个时间复杂度比上一个解法好一些，避免了不必要的比较。

1:    string longestCommonPrefix(vector<string> &strs) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      string prefix;
5:      if(strs.size() ==0) return prefix;
6:      int k=0;
7:      while(1)
8:      {
9:        if(k == strs[0].size()) break;
10:        char p = strs[0][k];
11:        int i =1;
12:        for(; i< strs.size(); i++)
13:        {
14:          if(k==strs[i].size()) break;
15:          if(p != strs[i][k])
16:            break;
17:        }
18:        if(i != strs.size())
19:          break;
20:        prefix.append(1,p);
21:        k++;
22:      }
23:      return prefix;
24:    }

Update 2, 3/7/2013
refactor code a bit

1:     string longestCommonPrefix(vector<string> &strs) {
2:            string prefix;
3:            if(strs.size() ==0) return prefix;
4:            int len =0;
5:            while(1)
6:            {
7:                 char var;
8:                 int i=0;
9:                 for(; i< strs.size(); i++)
10:                 {
11:                      if(i ==0) var =strs[0][len];
12:                      if(strs[i].size() == len || var != strs[i][len])
13:                      break;
14:                 }
15:                 if(i!= strs.size())
16:                      break;
17:                 len++;
18:                 prefix.append(1, var);
19:            }
20:            return prefix;
21:       }