Solution of 0003 and 0004 in c language and readme done

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

@@ -300,6 +300,34 @@ class Solution {
 }
 ```
 
+#### C
+
+```c
+int lengthOfLongestSubstring(char *s) {
+  int freq[256] = {0};
+  int l = 0, r = 0;
+  int ans = 0;
+  int len = strlen(s);
+
+  for (r = 0; r < len; r++) {
+    char c = s[r];
+    freq[(unsigned char)c]++;
+
+    while (freq[(unsigned char)c] > 1) {
+      freq[(unsigned char)s[l]]--;
+      l++;
+    }
+
+    if (ans < r - l + 1) {
+      ans = r - l + 1;
+    }
+  }
+
+  return ans;
+}
+
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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

@@ -298,6 +298,35 @@ class Solution {
 }
 ```
 
+#### C
+
+```c
+
+int lengthOfLongestSubstring(char *s) {
+  int freq[256] = {0};
+  int l = 0, r = 0;
+  int ans = 0;
+  int len = strlen(s);
+
+  for (r = 0; r < len; r++) {
+    char c = s[r];
+    freq[(unsigned char)c]++;
+
+    while (freq[(unsigned char)c] > 1) {
+      freq[(unsigned char)s[l]]--;
+      l++;
+    }
+
+    if (ans < r - l + 1) {
+      ans = r - l + 1;
+    }
+  }
+
+  return ans;
+}
+
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0000-0099/0003.Longest Substring Without Repeating Characters/Solution.c

@@ -0,0 +1,22 @@
+int lengthOfLongestSubstring(char *s) {
+  int freq[256] = {0};
+  int l = 0, r = 0;
+  int ans = 0;
+  int len = strlen(s);
+
+  for (r = 0; r < len; r++) {
+    char c = s[r];
+    freq[(unsigned char)c]++;
+
+    while (freq[(unsigned char)c] > 1) {
+      freq[(unsigned char)s[l]]--;
+      l++;
+    }
+
+    if (ans < r - l + 1) {
+      ans = r - l + 1;
+    }
+  }
+
+  return ans;
+}

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

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

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

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

solution/0000-0099/0004.Median of Two Sorted Arrays/Solution.c

@@ -0,0 +1,25 @@
+int findKth(int *nums1, int m, int i, int *nums2, int n, int j, int k) {
+  if (i >= m)
+    return nums2[j + k - 1];
+  if (j >= n)
+    return nums1[i + k - 1];
+  if (k == 1)
+    return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
+
+  int p = k / 2;
+
+  int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
+  int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
+
+  if (x < y)
+    return findKth(nums1, m, i + p, nums2, n, j, k - p);
+  else
+    return findKth(nums1, m, i, nums2, n, j + p, k - p);
+}
+
+double findMedianSortedArrays(int *nums1, int m, int *nums2, int n) {
+  int total = m + n;
+  int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
+  int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
+  return (a + b) / 2.0;
+}