[Leetcode] 28, 8, 5

28. Implement strStr()

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Solution(1): 直接暴力求解，时间复杂度O(m*n)，面试时这样做就可以了。

Code:

class Solution {public:    int strStr(string haystack, string needle) {        if(needle.size()==0) return 0;        if(haystack.size() < needle.size()) return -1;        for(int i=0; i<=haystack.size()-needle.size(); i++){            int t;            for(t=0; t<needle.size(); t++){                if(needle[t] != haystack[i+t]) break;            }            if(t==needle.size()) return i;        }        return -1;    }};

Solution(2): 使用KMP算法快速求解，时间复杂度O(m+n)。

注意：如果使用vector<int> next(s.size(),0); 在函数返回时会报错

首先求出needle对应的next数组：

然后在求subString的时候，遇到相异点时，可以通过next数组快速找到子序列中可以复用的部分：

详细解释参考博客：

http://www.cnblogs.com/c-cloud/p/3224788.html

http://blog.csdn.net/yutianzuijin/article/details/11954939/

Code:

class Solution {public:    int strStr(string haystack, string needle) {        if(haystack.size()<needle.size()) return -1;        else if(needle.size()==0) return 0;        else if(needle.size()==1){            for(int i=0; i<haystack.size(); i++){                if(haystack[i]==needle[0]) return i;            }            return -1;        }                        vector<int> next;        for(int i=0; i<needle.size(); i++)            next.push_back(0);        compute_next(needle,next);        int ih = 0; //haystack下标        int in = 0; //needle下标        while(ih<haystack.size() && in<needle.size()){            if(haystack[ih]==needle[in]){                ih++;                in++;                                if(in==needle.size()){                    return ih-in;                }            }else if(in!=0){                //in=3时，表示前面已有3个字符相等，因此找3个字符组成的子序列中的前后缀最长重合部分个数next[3]                //此时haystack起点处为ih-in                in = next[in];            }else{                ih++;            }        }        cout<<"end"<<endl;        return -1;    }private:    static void compute_next(string needle, vector<int>& next){        // needle.size > 1        //（动规）用于计算KMP算法中的next数组        // next[i]表示needle前i位构成的子数组的最长相等前缀后缀的长度        // needle = abcdefabc... -> next[8] = 3        next[0] = 0;        next[1] = 0;        for(int i=1; i<needle.size(); i++){            //next[i]->needle[i-1]，注意两个数组下标是错开的            if(needle[i]==needle[next[i]]){                next[i+1] = next[i]+1;            }else{                next[i+1] = 0;                int index = next[i];                do{                    index = next[index];                    if(needle[i]==needle[index]){                        next[i+1] = index+1;                        break;                    }                }while(index>0);            }        }    }};

8. String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

Solution: 此题不难，只要注意处理几种特殊情况即可。

Code:

class Solution {public:    int myAtoi(string str) {        int ans = 0;        if(str.size()==0) return 0;                int i=0;                //注意空格，空格在最前面时合法        for(;i<str.size(); i++)            if(str[i]!=' ') break;                //注意正负号，正负号直接接着数字合法        int sign = 1;        if(str[i]=='+') i++;        else if(str[i]=='-'){            sign = -1;            i++;        }                        for(; i<str.size(); i++){            if(str[i]>'9' || str[i]<'0') break; //遇到非数字停止            if(ans > INT_MAX/10 ||                (ans == INT_MAX/10 && str[i]-'0' > INT_MAX % 10)) //注意数字越界，越界就不再累积                return sign==-1 ? INT_MIN:INT_MAX;            ans = ans*10 + (str[i]-'0');                        }        return sign * ans;    }};

5. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"Output: "bab"Note: "aba" is also a valid answer.

Example:

Input: "cbbd"Output: "bb"

Solution(1):直接暴力求解即可，O(n^3)。

Code:

class Solution {public:    string longestPalindrome(string s) {        string pstring = "";        for(int i=0; i<s.size(); i++){            string newstring = "";            newstring.push_back(s[i]);            for(int t=s.size()-1; t>i; t--){                int begin = i;                int end = t;                while(begin<end){                    if(s[begin]!=s[end]) break;                    begin++;                    end--;                }                if(begin>=end){                    newstring = s.substr(i,t-i+1);                    break;                }            }            if(newstring.size()>pstring.size())                pstring = newstring;        }        return pstring;    }};

Solution(2): 暴力求解2，从中间向两边找，即可O(n^2)。

Code:

class Solution {public:    string longestPalindrome(string s) {        string pstring = "";        for(int i=0; i<s.size(); i++){            //以i作为中间的单独数            int t0 = i-1;            int t1 = i+1;            int count = 1;            calsubPalin(t0,t1,count,s);            if(count>pstring.size()) pstring = s.substr(t0,t1-t0+1);            //以i和i-1或i和i+1作为中间的数            count = 0;            t0 = i-1;            t1 = i;            calsubPalin(t0,t1,count,s);            if(count>pstring.size()) pstring = s.substr(t0,t1-t0+1);            count = 0;            t0 = i;            t1 = i+1;            calsubPalin(t0,t1,count,s);            if(count>pstring.size()) pstring = s.substr(t0,t1-t0+1);        }        return pstring;    }private:    static void calsubPalin(int& t0, int& t1, int& count, string& s){                while(t0>=0 && t1<s.size()){            if(s[t0]==s[t1]){                count+=2;                t0--;                t1++;            }            else break;        }        t0++;        t1--;    }};

Solution(3): 经典算法Manacher’s Algorithm，时间复杂度O(n)。

详细解释参考：http://blog.csdn.net/hopeztm/article/details/7932245

注意：这篇博客有个地方是错的。if P[ i' ] ≤ R – i 改成 if P[ i' ] ＜ R – i ，否则会出错，因为当半径正好到边界的时候，也要进一步试探真实长度。

Code:

class Solution {public:    string longestPalindrome(string s) {        string dummys = "$";//避免试探R范围时需要检查边界        //插入'#'，避免分类讨论        for(int i=0; i<s.size(); i++){            dummys.push_back('#');            dummys.push_back(s[i]);        }        dummys.push_back('#');                vector<int> P;        P.push_back(0);        P.push_back(0);        int C = 1;         int maxC = 1;        for(int i=2; dummys[i]; i++){            if(i < C+P[C]){                int t = 2*C-i;                if(P[t]<C+P[C]-i) P.push_back(P[t]);                else{                    int R = C+P[C]-i;                    //试探R的范围                    while(dummys[i-R] == dummys[i+R]) R++;                    R--;                    P.push_back(R);                }                if(P[i]+i>P[C]+C) C = i;                if(P[i]>P[maxC]) maxC = i;            }else{                int R = 1;                //试探R的范围                while(dummys[i-R] == dummys[i+R]) R++;                R--;                P.push_back(R);                if(P[i]+i>P[C]+C) C = i;                if(P[i]>P[maxC]) maxC = i;            }        }                string ans = "";        for(int i=maxC-P[maxC]; i<=maxC+P[maxC]; i++){            if(dummys[i]!='#') ans.push_back(dummys[i]);        }        return ans;    }};