LeetCode 22 Generate Parentheses (C,C++,Java,Python)
阿新 • • 發佈:2019-02-11
Problem:
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
Solution:
遞迴遍歷即可,注意兩個條件,要插入‘(’的時候一定要剩餘的數目大於0,要插入‘)’的時候一定要剩餘的數目大於0 並且剩餘的)的數目大於剩餘的( 的數目題目大意:
Java原始碼(252ms):
public class Solution { public List<String> generateParenthesis(int n) { List<String> res = new ArrayList<String>(); char[] tmp=new char[n+n]; generate(res,n,n,tmp,0); return res; } private void generate(List<String> res,int l,int r,char[] tmp,int index){ if(l==0 && r==0){ res.add(new String(tmp)); return; } if(l>0){ tmp[index]='('; generate(res,l-1,r,tmp,index+1); } if(r>0 && r>l){ tmp[index]=')'; generate(res,l,r-1,tmp,index+1); } } }
C語言原始碼(2ms):
void generate(char** res,int* size,int l,int r,char* tmp,int index){ if(l==0 && r==0){ tmp[index]=0; res[*size]=(char*)malloc(sizeof(char)*index); strcpy(res[*size],tmp); (*size)++; return; } if(l>0){ tmp[index]='('; generate(res,size,l-1,r,tmp,index+1); } if(r>0 && l<r){ tmp[index]=')'; generate(res,size,l,r-1,tmp,index+1); } } char** generateParenthesis(int n, int* returnSize) { char** res; char* tmp=(char*)malloc(sizeof(char)*(n*2+1)); int l=n,r=n; res=(char**)malloc(sizeof(char*)*1000000); *returnSize=0; generate(res,returnSize,l,r,tmp,0); return res; }
C++原始碼(4ms):
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> res;
char* tmp=(char*)malloc(sizeof(char)*(n+n+1));
generate(res,n,n,tmp,0);
return res;
}
private:
void generate(vector<string>& res,int l,int r,char* tmp,int index){
if(l==0 && r==0){
tmp[index]=0;
res.push_back(string(tmp));
return;
}
if(l>0){
tmp[index]='(';
generate(res,l-1,r,tmp,index+1);
}
if(r>0 && r>l){
tmp[index]=')';
generate(res,l,r-1,tmp,index+1);
}
}
};
Python原始碼(57ms):
class Solution:
# @param {integer} n
# @return {string[]}
def generateParenthesis(self, n):
res=[]
tmp=['' for i in range(n+n)]
self.generate(res,n,n,tmp,0)
return res
def generate(self,res,l,r,tmp,index):
if l==0 and r==0:
res.append(''.join(tmp))
return
if l>0:
tmp[index]='('
self.generate(res,l-1,r,tmp,index+1)
if r>0 and r>l:
tmp[index]=')'
self.generate(res,l,r-1,tmp,index+1)