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-rw-r--r--3sum.c77
-rw-r--r--3sum.py47
2 files changed, 124 insertions, 0 deletions
diff --git a/3sum.c b/3sum.c
new file mode 100644
index 0000000..40bed54
--- /dev/null
+++ b/3sum.c
@@ -0,0 +1,77 @@
+// 15. 3Sum
+#include "leetcode.h"
+
+/*
+ * given an integer array 'nums', return all the triplets '[nums[i], nums[j],
+ * nums[k]]' such that 'i != j', 'i != k', 'j != k', and 'nums[i] + nums[j] +
+ * nums[k] == 0'. notice that the solution set must not contain duplicate
+ * triplets
+ */
+
+int cmp(const void *a, const void *b) { return *((int *)a) - *((int *)b); }
+
+int **threeSum(int *nums, int numsSize, int *returnSize,
+ int **returnColumnSizes) {
+ qsort(nums, numsSize, sizeof(int), cmp);
+ (*returnSize) = 0;
+ (*returnColumnSizes) = (int *)malloc(sizeof(int) * numsSize * numsSize);
+ int **ans = (int **)malloc(numsSize * numsSize * sizeof(int *));
+ for (int i = 0; i < numsSize - 2; i++)
+ if (!i || nums[i] != nums[i - 1]) {
+ int l = i + 1, r = numsSize - 1;
+ while (l < r) {
+ if (nums[i] + nums[l] + nums[r] < 0)
+ l++;
+ else if (nums[i] + nums[l] + nums[r] > 0)
+ r--;
+ else {
+ ans[(*returnSize)] = (int *)malloc(3 * sizeof(int));
+ (*returnColumnSizes)[(*returnSize)] = 3;
+ ans[(*returnSize)][0] = nums[i];
+ ans[(*returnSize)][1] = nums[l];
+ ans[(*returnSize)][2] = nums[r];
+ (*returnSize)++;
+ l++;
+ while (l < r && nums[l] == nums[l - 1])
+ l++;
+ }
+ }
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {-1, 0, 1, 2, -1, -4}, rs1, *rcs1;
+ int n2[] = {0, 1, 1}, rs2, *rcs2;
+ int n3[] = {0, 0, 0}, rs3, *rcs3;
+ int **ts1 = threeSum(n1, ARRAY_SIZE(n1), &rs1, &rcs1);
+ int **ts2 = threeSum(n2, ARRAY_SIZE(n2), &rs2, &rcs2);
+ int **ts3 = threeSum(n3, ARRAY_SIZE(n3), &rs3, &rcs3);
+ for (int i = 0; i < rs1; i++) {
+ for (int j = 0; j < *rcs1; j++)
+ printf("%d ", ts1[i][j]); // expect: [[-1,-1,2],[-1,0,1]]
+ printf("\n");
+ }
+ printf("\n");
+ for (int i = 0; i < rs2; i++) {
+ for (int j = 0; j < *rcs2; j++)
+ printf("%d ", ts2[i][j]); // expect: null
+ printf("\n");
+ }
+ printf("\n");
+ for (int i = 0; i < rs3; i++) {
+ for (int j = 0; j < *rcs3; j++)
+ printf("%d ", ts3[i][j]); // expect: [[0,0,0]]
+ printf("\n");
+ }
+ printf("\n");
+ for (int i = 0; i < rs1; i++)
+ free(ts1[i]);
+ free(ts1);
+ for (int i = 0; i < rs2; i++)
+ free(ts2[i]);
+ free(ts2);
+ for (int i = 0; i < rs3; i++)
+ free(ts3[i]);
+ free(ts3);
+}
diff --git a/3sum.py b/3sum.py
new file mode 100644
index 0000000..9ac89e1
--- /dev/null
+++ b/3sum.py
@@ -0,0 +1,47 @@
+# 15. 3Sum
+
+"""
+given an integer array 'nums', return all the triplets '[nums[i], nums[j],
+nums[k]]' such that 'i != j', 'i != k', 'j != k', and 'nums[i] + nums[j] +
+nums[k] == 0'. notice that the solution set must not contain duplicate
+triplets
+"""
+
+
+class Solution:
+ def threeSum(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: List[List[int]]
+ """
+ nums.sort()
+ ans = []
+ if len(nums) < 3:
+ return ans
+ for i in range(len(nums)):
+ if nums[i] > 0:
+ break
+ if i > 0 and nums[i] == nums[i - 1]:
+ continue
+ j, k = i + 1, len(nums) - 1
+ while j < k:
+ total_sum = nums[i] + nums[j] + nums[k]
+ if total_sum > 0:
+ k -= 1
+ elif total_sum < 0:
+ j += 1
+ else:
+ ans.append([nums[i], nums[j], nums[k]])
+ prev_j, prev_k = nums[j], nums[k]
+ while j < k and nums[j] == prev_j:
+ j += 1
+ while j < k and nums[k] == prev_k:
+ k -= 1
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.threeSum(nums=[-1, 0, 1, 2, -1, -4]))
+ print(obj.threeSum(nums=[0, 1, 1]))
+ print(obj.threeSum(nums=[0, 0, 0]))