Leetcode Practice-String

First, palindrome-partitioning

Topic description
Given a string s, partition s such that every substring to the partition is a palindrome.
Return all possible palindrome partitioning of S.
For example, given s = "AaB",
Return
[
["AA", "B"],
["A", "a", "B"]
]

Train of thought: Dynamic programming + depth First traversal (DFS)

1, using the dynamic specification algorithm to divide the sub-palindrome string

2. Using DFS to traverse possible situations

Code:

Class Solution {public:vector<vector<string> > partition (String s) {VECTOR&LT;VECTOR&LT;STRING&G T
        > result;
        Vector<string> Array;
        if (S.length () ==0) return result;

        vector<vector<bool> > Flag = DP (s);
        DFS (s, 0, flag, array, result);
    return result; } void Dfs (string s, int begin, Vector<vector<bool> > Flag, vector<string> Array, vector<vector& 
            lt;string> > &result) {if (Begin==s.length ()) {result.push_back (array);
            Array.clear ();  
        Return for (int i=begin;i<s.length (); ++i) {if (flag[begin][i]==1) {Vector<stri
                Ng> Temp (array);
                Temp.push_back (S.substr (Begin,i+1-begin)); cout << "begin=" << begin << ";
                I+1= "<< (i+1) << Endl; cout << array.back () << EndL  
            DFS (s, i+1, flag, temp, result);
        }} vector<vector<bool> > dp (string s) {int len = s.length ();
        vector<vector<bool> > Flag (len, vector<bool> (len, 0));
        int I, J;  for (i=len-1; i>=0;-i) {for (j=i; j<len; ++j) {if (j = = i) flag[i][j]
                = true; else{if (s[i] = = S[j] && (j = = I+1 | | flag[i+1][j-1] = = True)) {Flag[i
                    ][J] = true;
    }}} return flag; }

};

Local Test code:

Vector<vector<bool> > DP (string s);//dynamic programming, molecular string
void Dfs (string s, int begin, Vector<vector<bool > > Flag, vector<string> array, vector<vector<string> > &result);//Depth First traversal
vector< Vector<string> > partition (string s);
int main () {
	string s = ' Abcbad ';
	cout << s.substr (1, 2) << Endl;

	vector<vector<bool> > Flag = DP (s);
	int I, J;
	For (i=0 i<flag.size (); ++i) {//Print flag for
		(j=0; j<flag[i].size (); ++j)
			cout << Flag[i][j] << ' ';
		cout << Endl;
	}

	vector<vector<string> > result;
	vector<string> Array;
	DFS (s, 0, flag, array, result);
	vector<vector<string> > result = partition (s);
	
	For (i=0 i<result.size (), ++i) {for
		(j=0; j<result[i].size (); ++j) {
			cout << Result[i][j] < < ';
		}
		cout << Endl;
	}
	return 0;
}

Second, Add-binary

Topic description
Given two binary strings, return their sum (also a binary string).
For example,
A = "11"
b = "1"
Return "100"

Code:

Class Solution {public:string addbinary (string A, string b) {string sum;//output int up=0;//rounding
        int I=a.length ()-1;
        int J=b.length ()-1;
                while (i>=0 && j>=0) {int temp = a[i]+b[j]-' 0 '-' 0 ' + up;//the number on the current bit and if (temp = = 3) {
                up = 1;
            Sum.insert (Sum.begin (), ' 1 ');
                else if (temp = = 2) {up = 1;
            Sum.insert (Sum.begin (), ' 0 ');
                else if (temp = = 1) {up = 0;
            Sum.insert (Sum.begin (), ' 1 ');
                else if (temp = = 0) {up = 0;
            Sum.insert (Sum.begin (), ' 0 ');
            } I.
        --j; 
            while (i>=0) {//At this point if i>=0, description a long, operation of the remaining digits of a int temp = a[i]-' 0 ' +up;
                if (temp = = 2) {up = 1;
            Sum.insert (Sum.begin (), ' 0 '); else if (temp = 1) {up = 0;
            Sum.insert (Sum.begin (), ' 1 ');
                else if (temp = = 0) {up = 0;
            Sum.insert (Sum.begin (), ' 0 ');
        } I.
            while (j>=0) {//At this point if j>=0, description b length, the remaining number of digits to b operation int temp = b[j]-' 0 ' + up;
                if (temp = = 2) {up = 1;
            Sum.insert (Sum.begin (), ' 0 ');
                else if (temp = = 1) {up = 0;
            Sum.insert (Sum.begin (), ' 1 ');
                else if (temp = = 0) {up = 0;
            Sum.insert (Sum.begin (), ' 0 ');
        }--j;
        The IF (up = = 1)//Determines whether the highest bit has a carry Sum.insert (Sum.begin (), ' 1 ');
    return sum; }
};

Local Test code:

int main () {
	string A;
	string B;
	
	cout << "A:";
	Getline (Cin, a);
	cout << "B:";
	Getline (CIN, b);
	
	string s = Addbinary (A, b);
	string s = "ss";
	cout << s << endl;
	return 0;
}