leetcode (22) - Generate Parentheses
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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:
[ "((()))", "(()())", "(())()", "()(())", "()()()"]
/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */void generate(char** res,int* size,int l,int r,char* tmp,int index){ // l和r记录剩余左右括号的数量 if(l==0 && r==0){ // 当且仅当左右括号数量为0时,正常结束 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); //每一层递归只有这一行函数结束了,才进行下面的if(r>0 && l<r)。再考虑最外层,也是如此。 } if(r>0 && l<r){ //剩余右括号数量比左括号多时,才能添加右括号 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;}
凑着这篇文章,理解一下递归:
递归的结束条件: 栈顶,一层一层的往回返,一直返回到最外层
递归的过程: 压栈,参数变化,参数可看作全局变量
每一层递归: 改变参数
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