Update Solution.c

Author: yanglbme

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

@@ -1,25 +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 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 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;
+    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);
+    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;
-}
+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;
+}