题记:
这道题不难但是很有意思,有两种解题思路,可以说一种是横向扫描,一种是纵向扫描。
横向扫描:遍历所有字符串,每次跟当前得出的最长公共前缀串进行对比,不断修正,最后得出最长公共前缀串。
纵向扫描:对所有串,从字符串第0位开始比较,全部相等则继续比较第1,2...n位,直到发生不全部相等的情况,则得出最长公共前缀串。
横向扫描算法实现:
//LeetCode_Longest Common Prefix
//Written by zhou
//2013.11.22 class Solution {
public:
string longestCommonPrefix(vector<string> &strs) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case. if (strs.size() == )
return ""; string prefix = strs[];
for (int i = ; i < strs.size(); ++i)
{
if (prefix.length() == || strs[i].length() == )
return ""; int len = prefix.length() < strs[i].length() ? prefix.length() : strs[i].length(); int j; for (j = ; j < len; ++j)
{
if (prefix[j] != strs[i][j])
break;
} prefix = prefix.substr(,j); } return prefix;
}
};
纵向扫描代码实现:
function longestCommonPrefix(strs:Array):String{
if(strs == null || strs.length == 0)return; for(i:int = 0; i < str[0].length - 1; i++)
{
for(j:int = 0; j < str.length - 1; j++)
{
if(str[i].charAt(j) != str[0].charAt(j))return strs[0].substr(0,j);
} }
return strs[0];
}