题目：
题解：
#define ADDITION -1
#define SUBTRACTION -2
#define MULTIPLICATION -3int* diffWaysToCompute(char * expression, int* returnSize) {int len = strlen(expression);int *ops = (int *)malloc(sizeof(int) * len);int opsSize = 0;for (int i = 0; i < len;) {if (!isdigit(expression[i])) {if (expression[i] == '+') {ops[opsSize++] = ADDITION;} else if (expression[i] == '-') {ops[opsSize++] = SUBTRACTION;} else {ops[opsSize++] = MULTIPLICATION;}i++;} else {int t = 0;while (i < len && isdigit(expression[i])) {t = t * 10 + expression[i] - '0';i++;}ops[opsSize++] = t;}}struct ListNode ***dp = NULL;dp = (struct ListNode ***)malloc(sizeof(struct ListNode **) * opsSize);for (int i = 0; i < opsSize; i++) {dp[i] = (struct ListNode **)malloc(sizeof(struct ListNode *) * opsSize);for (int j = 0; j < opsSize; j++) {dp[i][j] = NULL;}}for (int i = 0; i < opsSize; i += 2) {struct ListNode *node = (struct ListNode*)malloc(sizeof(struct ListNode));node->val = ops[i];node->next = NULL;dp[i][i] = node;}for (int i = 3; i <= opsSize; i++) {for (int j = 0; j + i <= opsSize; j += 2) {int l = j;int r = j + i - 1;for (int k = j + 1; k < r; k += 2) {struct ListNode *left = dp[l][k - 1];struct ListNode *right = dp[k + 1][r];for (struct ListNode *left = dp[l][k - 1]; left; left = left->next) {for (struct ListNode *right = dp[k + 1][r]; right; right = right->next) {struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode));if (ops[k] == ADDITION) {node->val = left->val + right->val;}else if (ops[k] == SUBTRACTION) {node->val = left->val - right->val;}else if (ops[k] == MULTIPLICATION) {node->val = left->val * right->val;}node->next = dp[l][r];dp[l][r] = node;}}}}}int *ans = (int *)malloc(sizeof(int) * (1 << opsSize));int pos = 0;for (struct ListNode *node = dp[0][opsSize - 1]; node; node = node->next) {ans[pos++] = node->val;}*returnSize = pos;for (int i = 0; i < opsSize; i++) {for (int j = 0; j < opsSize; j++) {struct ListNode *curr, *tmp;curr = dp[i][j];while (curr) {tmp = curr;curr = curr->next;free(tmp);}}free(dp[i]);}free(dp);free(ops);return ans;
}
