Approach

# 枚舉各個長度的字串

Time Complexity

O(n * n)

Space Complexity

O(1)

Code

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        string p1 = strs[0];

        for (int i = 1 ; i < strs.size() ; i++) {
            while (strs[i].find(p1) != 0) {
                p1 = p1.substr(0 , p1.length() - 1);

                if (p1.empty()) {
                    return "";
                }
            }
        }

        return p1;
    }
};