*[topcoder]BracketExpressions
http://community.topcoder.com/stat?c=problem_statement&pm=13243
就是能否通过把字符串中的'X'替换成"()", "[]", and "{}"来变成合法的括号字符串，
"([]X()[()]XX}[])X{{}}]"
Returns: "possible"
You can replace 'X's respectively with '{', '(', ')' and '['.
DFS搜索，Valid的判断使用stack。
#include <stack>
#include <vector>
#include <string>
using namespace std;
class BracketExpressions {
public:
string candidates;
string ifPossible(string expression)
{
candidates = "()[]{}";
vector<int> xPos;
for (int i = 0; i < expression.length(); i++)
{
if (expression[i] == 'X')
{
xPos.push_back(i);
}
}
bool possible = ifPossRe(expression, 0, xPos);
if (possible)
return "possible";
else
return "impossible";
}
bool isValid(const string &expression)
{
stack<char> stk;
for (int i = 0; i < expression.length(); i++)
{
if (stk.empty())
{
stk.push(expression[i]);
}
else if (match(stk.top(), expression[i]))
{
stk.pop();
}
else
{
stk.push(expression[i]);
}
}
return stk.empty();
}
bool match(char a, char b)
{
if (a == '(' && b == ')') return true;
if (b == '(' && a == ')') return true;
if (a == '[' && b == ']') return true;
if (b == '[' && a == ']') return true;
if (a == '{' && b == '}') return true;
if (b == '{' && a == '}') return true;
return false;
}
bool ifPossRe(string &expression, int idx, vector<int> &xPos)
{
if (idx == xPos.size())
{
return isValid(expression);
}
int n = xPos[idx];
for (int i = 0; i < candidates.length(); i++)
{
char ch = expression[n];
expression[n] = candidates[i];
bool res = ifPossRe(expression, idx+1, xPos);
expression[n] = ch;
if (res)
return true;
}
return false;
}
};
http://apps.topcoder.com/wiki/display/tc/SRM+628
但其实判断Backet合法的代码是错的，因为没有判断先有左括号再有右括号，以下做法更好更简洁。
bool correctBracket(string exp)
{
stack<char> s;
// an assoicaitive array: opos[ ')' ] returns '(', opos[ ']' ] is '[', ...
map<char,char> opos = {
{ ')', '(' },
{ ']', '[' },
{ '}', '{' },
};
for (char ch: exp) {
// we push opening brackets to the stack
if (ch == '(' || ch == '[' || ch == '{' ) {
s.push(ch);
} else {
// If we find a closing bracket, we make sure it matches the
// opening bracket in the top of the stack
if (s.size() == 0 || s.top() != opos[ch]) {
return false;
} else {
// then we remove it
s.pop();
}
}
}
// stack must be empty.
return s.empty();
}
解法中还是用了基于6的幂来计算所有组合，比DFS要快。
string ifPossible(string expression)
{
vector<int> x;
int n = expression.size();
for (int i = 0; i < n; i++) {
if (expression[i] == 'X') {
x.push_back(i);
}
}
int t = x.size();
// to easily convert to base 6 we precalculate the powers of 6:
int p6[6];
p6[0] = 1;
for (int i = 1; i < 6; i++) {
p6[i] = 6 * p6[i - 1];
}
const char* CHARS = "([{)]}";
for (int m = 0; m < p6[t]; m++) {
string nexp = expression;
for (int i = 0; i < t; i++) {
// (m / p6[i]) % 6 extracts the i-th digit of m in base 6.
nexp[ x[i] ] = CHARS[ (m / p6[i]) % 6 ];
}
if (correctBracket(nexp)) {
return "possible";
}
}
return "impossible";
}