LeetCode 14 Longest Common Prefix
2017-01-09 15:43
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题意:
最长公共前缀
思路:
一个字符一个字符的考虑即可,但是写代码时候要小心,也许当前string已经不够长了,或者它一个字符串都没给你…
官方题解很成功的把简单题想得巨烦,建议看着玩玩即可,思考下大数据下算法的扩展性。
代码:
//
// Created by house on 1/9/17.
//
class Solution {
public:
string longestCommonPrefix(vector <string> &strs) {
if (strs.size() == 0) {
return "";
}
if (strs.size() == 1) {
return strs[0];
}
int ans = 0;
for (int i = 1;; ++i) {
bool same = true;
if (strs[0].size() < i) {
break;
}
char y = strs[0][i - 1];
for (string x : strs) {
if (x.size() < i || x[i - 1] != y) {
same = false;
break;
}
}
if (!same) {
break;
}
ans = i;
}
if (ans) {
return strs[0].substr(0, ans);
}
return "";
}
};
最长公共前缀
思路:
一个字符一个字符的考虑即可,但是写代码时候要小心,也许当前string已经不够长了,或者它一个字符串都没给你…
官方题解很成功的把简单题想得巨烦,建议看着玩玩即可,思考下大数据下算法的扩展性。
代码:
//
// Created by house on 1/9/17.
//
class Solution {
public:
string longestCommonPrefix(vector <string> &strs) {
if (strs.size() == 0) {
return "";
}
if (strs.size() == 1) {
return strs[0];
}
int ans = 0;
for (int i = 1;; ++i) {
bool same = true;
if (strs[0].size() < i) {
break;
}
char y = strs[0][i - 1];
for (string x : strs) {
if (x.size() < i || x[i - 1] != y) {
same = false;
break;
}
}
if (!same) {
break;
}
ans = i;
}
if (ans) {
return strs[0].substr(0, ans);
}
return "";
}
};
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