LeetCode/最长公共前缀

1. 两两比较

把当前最长公共前缀与每一个字符串比较，并进行更新减小

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        string prefix = strs[0];
        int count = strs.size();
        for (int i = 1; i < count; ++i) {
            prefix = longestCommonPrefix(prefix, strs[i]);
            if (!prefix.size()) {
                break;
            }
        }
        return prefix;
    }

    string longestCommonPrefix(const string& str1, const string& str2) {
        int length = min(str1.size(), str2.size());
        int index = 0;
        while (index < length && str1[index] == str2[index]) {
            ++index;
        }
        return str1.substr(0, index);
    }
};

2. 全局比较

对每个字符串对应位置字符从前往后进行比较，最长公共前缀逐渐增大

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        int length = strs[0].size();
        int count = strs.size();
        for (int i = 0; i < length; ++i) {
            char c = strs[0][i];
            for (int j = 1; j < count; ++j) {
                if (i == strs[j].size() || strs[j][i] != c) {
                    return strs[0].substr(0, i);
                }
            }
        }
        return strs[0];
    }
};

自己写的

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        int i=-1;
        int maxlen=0;
        for(int j=0;j<strs.size();j++)
            if(strs[j].size()>maxlen)
             maxlen=strs[j].size();

        bool flag = true;
        char temp;
        while(flag){
            i++;
            if(i==maxlen) break;
            temp = strs[0][i];
            for(int j=0;j<strs.size();j++) if(strs[j][i]!=temp){flag=false;break;}
        }
        return strs[0].substr(0,i);
    }
};

3. 二分两两比较

递归分治，然后两两提取最长公共前缀，相当于二叉树形的两两比较

二分归并

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        else {
            return longestCommonPrefix(strs, 0, strs.size() - 1);//原函数不具有划分性质，重新写一个
        }
    }

    string longestCommonPrefix(const vector<string>& strs, int start, int end) {
        if (start == end) {
            return strs[start];
        }
        else {
            int mid = (start + end) / 2;
            string lcpLeft = longestCommonPrefix(strs, start, mid);//分治左
            string lcpRight = longestCommonPrefix(strs, mid + 1, end);//分治右
            return commonPrefix(lcpLeft, lcpRight);//合并分支结果
        }
    }

    string commonPrefix(const string& lcpLeft, const string& lcpRight) {//提取两字符串公共前缀
        int minLength = min(lcpLeft.size(), lcpRight.size());
        for (int i = 0; i < minLength; ++i) {
            if (lcpLeft[i] != lcpRight[i]) {
                return lcpLeft.substr(0, i);
            }
        }
        return lcpLeft.substr(0, minLength);
    }
};

4. 利用C++内置sort

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.empty()) return string();
        sort(strs.begin(), strs.end());
        string st = strs.front(), en = strs.back();
        int i, num = min(st.size(), en.size());
        for(i = 0; i < num && st[i] == en[i]; i ++);
        return string(st, 0, i);
    }
};