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Commit b2a6267

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feat: add c solutions to lc problems: No.0003,0004 (#4427)
1 parent 3e3d06c commit b2a6267

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6 files changed

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6 files changed

+161
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lines changed

‎solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md‎

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@@ -300,6 +300,33 @@ class Solution {
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}
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```
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#### C
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```c
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int lengthOfLongestSubstring(char* s) {
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int freq[256] = {0};
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int l = 0, r = 0;
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int ans = 0;
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int len = strlen(s);
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for (r = 0; r < len; r++) {
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char c = s[r];
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freq[(unsigned char) c]++;
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while (freq[(unsigned char) c] > 1) {
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freq[(unsigned char) s[l]]--;
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l++;
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}
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if (ans < r - l + 1) {
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ans = r - l + 1;
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/0000-0099/0003.Longest Substring Without Repeating Characters/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -298,6 +298,33 @@ class Solution {
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}
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```
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#### C
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```c
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int lengthOfLongestSubstring(char* s) {
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int freq[256] = {0};
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int l = 0, r = 0;
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int ans = 0;
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int len = strlen(s);
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for (r = 0; r < len; r++) {
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char c = s[r];
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freq[(unsigned char) c]++;
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while (freq[(unsigned char) c] > 1) {
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freq[(unsigned char) s[l]]--;
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l++;
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}
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if (ans < r - l + 1) {
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ans = r - l + 1;
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
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int lengthOfLongestSubstring(char* s) {
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int freq[256] = {0};
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int l = 0, r = 0;
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int ans = 0;
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int len = strlen(s);
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for (r = 0; r < len; r++) {
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char c = s[r];
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freq[(unsigned char) c]++;
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while (freq[(unsigned char) c] > 1) {
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freq[(unsigned char) s[l]]--;
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l++;
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}
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if (ans < r - l + 1) {
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ans = r - l + 1;
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}
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}
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return ans;
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}

‎solution/0000-0099/0004.Median of Two Sorted Arrays/README.md‎

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@@ -350,6 +350,36 @@ proc medianOfTwoSortedArrays(nums1: seq[int], nums2: seq[int]): float =
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# echo medianOfTwoSortedArrays(arrA, arrB)
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```
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#### C
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```c
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int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
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if (i >= m)
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return nums2[j + k - 1];
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if (j >= n)
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return nums1[i + k - 1];
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if (k == 1)
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return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
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int p = k / 2;
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int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
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int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
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if (x < y)
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return findKth(nums1, m, i + p, nums2, n, j, k - p);
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else
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return findKth(nums1, m, i, nums2, n, j + p, k - p);
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}
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double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
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int total = m + n;
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int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
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int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
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return (a + b) / 2.0;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/0000-0099/0004.Median of Two Sorted Arrays/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -346,6 +346,36 @@ proc medianOfTwoSortedArrays(nums1: seq[int], nums2: seq[int]): float =
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# echo medianOfTwoSortedArrays(arrA, arrB)
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```
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#### C
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```c
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int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
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if (i >= m)
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return nums2[j + k - 1];
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if (j >= n)
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return nums1[i + k - 1];
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if (k == 1)
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return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
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int p = k / 2;
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int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
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int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
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if (x < y)
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return findKth(nums1, m, i + p, nums2, n, j, k - p);
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else
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return findKth(nums1, m, i, nums2, n, j + p, k - p);
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}
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double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
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int total = m + n;
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int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
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int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
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return (a + b) / 2.0;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
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if (i >= m)
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return nums2[j + k - 1];
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if (j >= n)
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return nums1[i + k - 1];
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if (k == 1)
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return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
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int p = k / 2;
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int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
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int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
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if (x < y)
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return findKth(nums1, m, i + p, nums2, n, j, k - p);
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else
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return findKth(nums1, m, i, nums2, n, j + p, k - p);
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}
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double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
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int total = m + n;
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int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
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int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
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return (a + b) / 2.0;
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}

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