LeetCode_String_Easy

String Easy

Zigzag Conversion

The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   NA P L S I I GY   I   R

And then read line by line: “PAHNAPLSIIGYIR”

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.

class Solution {public:    string convert(string s, int numRows) {     int len=s.length();        string res;        if(numRows==1 || numRows>=len)        return s;        string ctmp="";        int itmp=0;        int count=0;        int tt=0;       for(int i=0;i<numRows;i++){            count=i;            itmp =1;            while(count<len){                ctmp=s.at(count);                res.append(ctmp);                if((i==0) || (i==(numRows-1))){                    tt=2*(numRows-1);                }                else{                    if(itmp%2==1)                        tt=2*((numRows-1)-i);                    else                         tt=2*i;                     itmp++;                }                count=count+tt;            }        }        return res;       }};

Add Binary

Given two binary strings, return their sum (also a binary string).

For example,
a = “11”
b = “1”
Return “100”.

class Solution {public:    string addBinary(string a, string b) {        string res;        char ts='0';        int lenA=a.length();        int lenB=b.length();        string tts="";        int diff=abs(lenA-lenB);        while(diff--){            if(lenA>lenB)                b.insert(0,"0");            else                a.insert(0,"0");        }        int len=max(lenA,lenB);        while(len>0){            len--;            if( a.at(len)!=b.at(len)){                if(ts=='0'){                    ts='0';                    res.insert(0,"1");                }else if(ts=='1'){                    ts='1';                    res.insert(0,"0");                }                       }            else if(a.at(len)==b.at(len) && ts=='1'){                if(a.at(len)=='0'){                    ts='0';                    res.insert(0,"1");                }                if(a.at(len)=='1'){                    ts='1';                    res.insert(0,"1");                }            }else if(a.at(len)==b.at(len) && ts=='0'){                if(a.at(len)=='0'){                    ts='0';                    res.insert(0,"0");                }                if(a.at(len)=='1'){                    ts='1';                    res.insert(0,"0");                }            }        }        if(ts=='1')            res.insert(0,"1");        return res;       }};

Valid Parentheses

Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.

The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.

class Solution {public:    bool isValid(string s) {        stack<char> st;        for (int i=0;i<s.size();i++){                                if ((s[i]=='(') ||(s[i]=='[') ||(s[i]=='{')) {st.push(s[i]);}            else{                   if (st.empty()){return false;}                if ((s[i]==')') && (st.top()!='(')) {return false;}                if ((s[i]=='{') && (st.top()!='}')) {return false;}                if ((s[i]=='[') && (st.top()!=']')) {return false;}                st.pop();            }        }        return st.empty();    }};