题目:
题解:
// 回溯法求解
#define MAX_SIZE 1430 // 卡特兰数: 1, 1, 2, 5, 14, 42, 132, 429, 1430
void generate(int left, int right, int n, char *str, int index, char **result, int *returnSize) {
if (index == 2 * n) { // 当前长度已达2n
result[(*returnSize)] = (char*)calloc((2 * n + 1), sizeof(char));
strcpy(result[(*returnSize)++], str);
return;
}
// 如果左括号数量不大于 n,可以放一个左括号
if (left < n) {
str[index] = '(';
generate(left + 1, right, n, str, index + 1, result, returnSize);
}
// 如果右括号数量小于左括号的数量,可以放一个右括号
if (right < left) {
str[index] = ')';
generate(left, right + 1, n, str, index + 1, result, returnSize);
}
}
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
char** generateParenthesis(int n, int *returnSize) {
char *str = (char*)calloc((2 * n + 1), sizeof(char));
char **result = (char **)malloc(sizeof(char *) * MAX_SIZE);
*returnSize = 0;
generate(0, 0, n, str, 0, result, returnSize);
return result;
}