Leetcode: Longest Common Prefix

class Solution
{
public:
    string longestCommonPrefix(vector<string> &strs)
    {
        if (strs.empty())
        {
            return "";
        }
        string common = strs[0];
        vector<string>::size_type size = strs.size();
        int length;//保存要比较的两个字符串的最小长度，只在最小长度范围内进行比较
        int count;//记录相等的字符个数
        for (int i = 1; i < size; i++)
        {
            length = min(common.length(), strs[i].length());
            count = 0;
            for (int j = 0; j < length; j++)
            {
                //如果两个字符相等count++
                if (strs[i][j] == common[j])
                {
                    count++;
                }
                //如果两个字符不相等直接退出内层循环
                else
                {
                    break;
                }
            }
            //将common和strs[i]的共同前缀保存在common中，进行下一个字符的比较
            common = common.substr(0, count);
        }
        return common;
    }
};

public class Solution
{
    public string LongestCommonPrefix(string[] strs)
    {
        if (strs == null || strs.Length == 0)
        {
            return string.Empty;
        }
        string common = strs[0];
        int length = 0;
        int count = 0;
        for (int i = 1; i < strs.Length; i++)
        {
            length = Math.Min(common.Length, strs[i].Length);
            count = 0;
            for (int j = 0; j < length; j++)
            {
                if (strs[i][j] == common[j])
                {
                    count++;
                }
                else
                {
                    break;
                }
            }
            common = common.Substring(0, count);
        }
        return common;
    }
}

class Solution:
    # @return a string
    def longestCommonPrefix(self, strs):
        size = len(strs)
        if not strs or size == 0:
            return ""
        common = strs[0]
        length = 0
        count = 0
        for i in range(1, size):
            length = min(len(common), len(strs[i]))
            count = 0
            for j in range(length):
                if strs[i][j] == common[j]:
                    count += 1
                else:
                    break
            common = common[0: count]
        return common