【简单】14. 最长公共前缀

vector.size

if (!strs.size()) {}
int count = strs.size();
if (!prefix.size()) {}

if (!strs.size()) {}
int length = strs[0].size();
int count = strs.size();
if (i == strs[j].size() || strs[j][i] != c) {}

if (!strs.size()) {}
return longestCommonPrefix(strs, 0, strs.size() - 1);
int minLength = min(lcpLeft.size(), lcpRight.size());

if (!strs.size()) {}
int count = strs.size();

std::vector::size public member function

Return size

Returns the number of elements in the vector.

This is the number of actual objects held in the vector, which is not necessarily equal to its storage capacity. alt

min

int length = min(str1.size(), str2.size());

std::min function template

Return the smallest

Returns the smallest of a and b. If both are equivalent, a is returned.

The versions for initializer lists (3) return the smallest of all the elements in the list. Returning the first of them if these are more than one.

The function uses operator< (or comp, if provided) to compare the values. alt

string.substr

return str1.substr(0, index);
  
return strs[0].substr(0, i);
  
return lcpLeft.substr(0, i);
return lcpLeft.substr(0, minLength);
  
return strs[0].substr(0, low);
string str0 = strs[0].substr(0, length);

std::string::substr public member function

  string substr (size_t pos = 0, size_t len = npos) const;

Generate substring

Returns a newly constructed string object with its value initialized to a copy of a substring of this object.

The substring is the portion of the object that starts at character position pos and spans len characters (or until the end of the string, whichever comes first). alt

14. 最长公共前缀

编写一个函数来查找字符串数组中的最长公共前缀。

如果不存在公共前缀，返回空字符串 ""。 alt

横向扫描

依次遍历字符串数组中的每个字符串，对于每个遍历到的字符串，更新最长公共前缀，当遍历完所有的字符串以后，即可得到字符串数组中的最长公共前缀。

如果在尚未遍历完所有的字符串时，最长公共前缀已经是空串，则最长公共前缀一定是空串，因此不需要继续遍历剩下的字符串，直接返回空串即可。

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);
    }
};

alt

纵向扫描

纵向扫描时，从前往后遍历所有字符串的每一列，比较相同列上的字符是否相同，如果相同则继续对下一列进行比较，如果不相同则当前列不再属于公共前缀，当前列之前的部分为最长公共前缀。

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {//如果不存在公共前缀，
            return "";//返回空字符串 ""
        }
        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];
    }
};

alt

分治

alt

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        else {
            return longestCommonPrefix(strs, 0, strs.size() - 1);
        }
    }

    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);
            string lcpRight = longestCommonPrefix(strs, mid + 1, end);
            return commonPrefix(lcpLeft, lcpRight);
        }
    }

    string commonPrefix(const string& lcpLeft, const string& lcpRight) {
        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);
    }
};

alt

二分查找

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if (!strs.size()) {
            return "";
        }
        int minLength = min_element(strs.begin(), strs.end(), [](const string& s, const string& t) {return s.size() < t.size();})->size();
        int low = 0, high = minLength;
        while (low < high) {
            int mid = (high - low + 1) / 2 + low;
            if (isCommonPrefix(strs, mid)) {
                low = mid;
            }
            else {
                high = mid - 1;
            }
        }
        return strs[0].substr(0, low);
    }

    bool isCommonPrefix(const vector<string>& strs, int length) {
        string str0 = strs[0].substr(0, length);
        int count = strs.size();
        for (int i = 1; i < count; ++i) {
            string str = strs[i];
            for (int j = 0; j < length; ++j) {
                if (str0[j] != str[j]) {
                    return false;
                }
            }
        }
        return true;
    }
};

alt