定义问题:左括号,右括号需要匹配,也就是必须相等
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
n = 3 ,说明需要3个左括号,3个右括号
left : 没有匹配的“(”
right:
dfs(int left, int right, char* str, char*result, int returnSize, int n)
left-1: 添加"("
right-1: 添加")"
二叉递归:[30,12,20,03,11,11,#,#,02,02,10,02,10]
recursion()
{
if (end_condition)
{
solve;
}
else
{
//在将问题转换为子问题描述的每一步,都解决该步中剩余部分的问题。
if()
recursion();
if()
recursion();
}
}
#define SIZE 10000
void dfs(int left, int right, char* str, char**result, int* returnSize, int n) {
if((left == 0) && (right == 0))
result[(*returnSize)++] = str;
else {
char* newStr = (char *)malloc(sizeof(char) * (2*n+1));
if(left > 0) {
strcpy(newStr, str);
dfs(left-1, right+1, strcat(newStr, "("), result, returnSize, n);
}
if(right > 0) {
strcpy(newStr, str);
dfs(left, right-1, strcat(newStr, ")"), result, returnSize, n);
}
}
}
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
char** generateParenthesis(int n, int* returnSize) {
char** result = (char **)malloc(sizeof(char *) * SIZE);
dfs(n, 0, "", result, returnSize, n);
return result;
}