diff options
-rw-r--r-- | diagonal_traverse.c | 57 | ||||
-rw-r--r-- | diagonal_traverse.py | 43 | ||||
-rw-r--r-- | longest_subarr_ones.c | 25 | ||||
-rw-r--r-- | longest_subarr_ones.py | 32 |
4 files changed, 145 insertions, 12 deletions
diff --git a/diagonal_traverse.c b/diagonal_traverse.c new file mode 100644 index 0000000..802271d --- /dev/null +++ b/diagonal_traverse.c @@ -0,0 +1,57 @@ +// 498. Diagonal Traverse +#include "leetcode.h" + +/* + * given an 'm x n' matrix 'mat', return an array of all the elements of the + * array in a diagonal order. + */ + +int *findDiagonalOrder(int **mat, int matSize, int *matColSize, + int *returnSize) { + int n = matSize * matColSize[0], row = 0, col = 0; + int *ans = (int *)malloc(n * sizeof(int)); + *returnSize = n; + for (int i = 0; i < n; i++) { + ans[i] = mat[row][col]; + if (!((row + col) % 2)) { + if (col == matSize - 1) + row++; + else if (!row) + col++; + else { + row--; + col++; + } + } else { + if (row == matColSize[0] - 1) + col++; + else if (!col) + row++; + else { + row++; + col--; + } + } + } + return ans; +} + +int main() { + int m1i[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, + m2i[2][2] = {{1, 2}, {3, 4}}; + struct two_d_arr m1, m2; + two_d_arr_init(&m1, ARRAY_SIZE(m1i), ARRAY_SIZE(m1i[0]), m1i); + two_d_arr_init(&m2, ARRAY_SIZE(m2i), ARRAY_SIZE(m2i[0]), m2i); + int rs1, *fdo1 = findDiagonalOrder(m1.arr, m1.row_size, m1.col_size, &rs1); + int rs2, *fdo2 = findDiagonalOrder(m2.arr, m2.row_size, m2.col_size, &rs2); + for (int i = 0; i < rs1; i++) + printf("%d ", fdo1[i]); // expect: 1 2 4 7 5 3 6 8 9 + printf("\n"); + for (int i = 0; i < rs2; i++) + printf("%d ", fdo2[i]); // expect: 1 2 3 4 + printf("\n"); + free(fdo1); + free(fdo2); + two_d_arr_free(&m1); + two_d_arr_free(&m2); +} diff --git a/diagonal_traverse.py b/diagonal_traverse.py new file mode 100644 index 0000000..d75deec --- /dev/null +++ b/diagonal_traverse.py @@ -0,0 +1,43 @@ +# 498. Diagonal Traverse + +""" +given an 'm x n' matrix 'mat', return an array of all the elements of the +array in a diagonal order. +""" + + +class Solution(object): + def findDiagonalOrder(self, mat): + """ + :type mat: List[List[int]] + :rtype: List[int] + """ + if not mat or not mat[0]: + return [] + m, n = len(mat), len(mat[0]) + ans, row, col = [], 0, 0 + for _ in range(m * n): + ans.append(mat[row][col]) + if (row + col) % 2 == 0: + if col == n - 1: + row += 1 + elif row == 0: + col += 1 + else: + row -= 1 + col += 1 + else: + if row == m - 1: + col += 1 + elif col == 0: + row += 1 + else: + row += 1 + col -= 1 + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.findDiagonalOrder(mat=[[1, 2, 3], [4, 5, 6], [7, 8, 9]])) + print(obj.findDiagonalOrder(mat=[[1, 2], [3, 4]])) diff --git a/longest_subarr_ones.c b/longest_subarr_ones.c index 049ce33..dc9aa84 100644 --- a/longest_subarr_ones.c +++ b/longest_subarr_ones.c @@ -1,6 +1,5 @@ // 1493. Longest Subarray of 1's After Deleting One Element -#include <math.h> -#include <stdio.h> +#include "leetcode.h" /* * given a binary array 'nums', you should delete one element from it. return @@ -8,24 +7,26 @@ * resulting array. return 0 if there is no such subarray. */ -int longestSubarray(int *nums, int nums_size) { - int n = nums_size, zero_cnt = 0, ans = 0, left = 0; - for (int right = 0; right < n; right++) { +int longestSubarray(int *nums, int numsSize) { + int left = 0, zeros = 0, ans = 0; + for (int right = 0; right < numsSize; right++) { if (!nums[right]) - zero_cnt++; - while (zero_cnt > 1) { + zeros++; + while (zeros > 1) { if (!nums[left]) - zero_cnt--; + zeros--; left++; } - ans = fmax(ans, right - left); + int length = right - left; + if (length > ans) + ans = length; } return ans; } int main() { int n1[] = {1, 1, 0, 1}, n2[] = {0, 1, 1, 1, 0, 1, 1, 0, 1}, n3[] = {1, 1, 1}; - printf("%d\n", longestSubarray(n1, 4)); // expect: 3 - printf("%d\n", longestSubarray(n2, 9)); // expect: 5 - printf("%d\n", longestSubarray(n3, 3)); // expect: 2 + printf("%d\n", longestSubarray(n1, ARRAY_SIZE(n1))); // expect: 3 + printf("%d\n", longestSubarray(n2, ARRAY_SIZE(n2))); // expect: 5 + printf("%d\n", longestSubarray(n3, ARRAY_SIZE(n3))); // expect: 2 } diff --git a/longest_subarr_ones.py b/longest_subarr_ones.py new file mode 100644 index 0000000..a4b3623 --- /dev/null +++ b/longest_subarr_ones.py @@ -0,0 +1,32 @@ +# 1493. Longest Subarray of 1's After Deleting One Element + +""" +given a binary array 'nums', you should delete one element from it. return +the size of the longest non-empty subarray containing only 1's in the +resulting array. return 0 if there is no such subarray. +""" + + +class Solution(object): + def longestSubarray(self, nums): + """ + :type nums: List[int] + :rtype: int + """ + left, zeros, ans = 0, 0, 0 + for right in range(len(nums)): + if nums[right] == 0: + zeros += 1 + while zeros > 1: + if nums[left] == 0: + zeros -= 1 + left += 1 + ans = max(ans, right - left) + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.longestSubarray(nums=[1, 1, 0, 1])) + print(obj.longestSubarray(nums=[0, 1, 1, 1, 0, 1, 1, 0, 1])) + print(obj.longestSubarray(nums=[1, 1, 1])) |