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-rw-r--r--product_array_except_self.c37
1 files changed, 37 insertions, 0 deletions
diff --git a/product_array_except_self.c b/product_array_except_self.c
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+// 238. Product of Array Except Self
+#include "leetcode.h"
+
+/*
+ * given an integer array 'nums' return an array 'answer' such that 'answer[i]'
+ * is equal to the product of all the elements of 'nums' except 'nums[i]'. the
+ * product of any prefix or suffix of 'nums' is guaranteed to fit in a 32-bit
+ * integer. you must write an algorithm that runs in O(n) time and without the
+ * division operation.
+ */
+
+int *productExceptSelf(int *nums, int nums_size, int *return_size) {
+ int *ans = (int *)malloc(nums_size * sizeof(int));
+ *return_size = nums_size;
+ ans[0] = 1;
+ for (int i = 1; i < nums_size; i++)
+ ans[i] = nums[i - 1] * ans[i - 1];
+ int tmp = 1;
+ for (int i = nums_size - 1; i >= 0; i--) {
+ ans[i] *= tmp;
+ tmp *= nums[i];
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {1, 2, 3, 4}, n2[] = {-1, 1, 0, -3, 3};
+ int rs1[] = {}, rs2[] = {};
+ int *pes1 = productExceptSelf(n1, ARRAY_SIZE(n1), rs1);
+ int *pes2 = productExceptSelf(n2, ARRAY_SIZE(n2), rs2);
+ for (int i = 0; i < ARRAY_SIZE(n1); i++)
+ printf("%d ", pes1[i]); // expect: 24 12 8 6
+ printf("\n");
+ for (int i = 0; i < ARRAY_SIZE(n2); i++)
+ printf("%d ", pes2[i]); // expect: 0 0 9 0 0
+ printf("\n");
+}