update

Author: OneCodeMonkey

leetcode_solved/leetcode_0088_Merge_Sorted_Array.cpp

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+/**
+ * AC: T:O(m+n) S:O(m+n)
+ * Runtime: 0 ms, faster than 100.00% of Java online submissions for Merge Sorted Array.
+ * Memory Usage: 38.8 MB, less than 89.83% of Java online submissions for Merge Sorted Array.
+ *
+ * 双指针法：可从头部开始，也可以从尾部开始
+ *
+ * 备选较差思路：合并后大数组排序，T:O((m+n)log(m+n)), S:O(log(m+n))
+ */
+/**
+ description:
+ Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
+
+The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has a size equal to m + n such that it has enough space to hold additional elements from nums2.
+
+
+Example 1:
+
+Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
+Output: [1,2,2,3,5,6]
+Example 2:
+
+Input: nums1 = [1], m = 1, nums2 = [], n = 0
+Output: [1]
+ 
+
+Constraints:
+
+nums1.length == m + n
+nums2.length == n
+0 <= m, n <= 200
+1 <= m + n <= 200
+-109 <= nums1[i], nums2[i] <= 109
+ *
+ */
+class Solution {
+    public static void merge(int[] nums1, int m, int[] nums2, int n) {
+        int p1 = 0, p2 = 0, retP = 0;
+        int[] ret = new int[m + n];
+        while (p1 != m && p2 != n) {
+            if (nums1[p1] < nums2[p2]) {
+                ret[retP] = nums1[p1];
+                p1++;
+            } else {
+                ret[retP] = nums2[p2];
+                p2++;
+            }
+            retP++;
+        }
+        if (p1 != m) {  // 数组1有尾部
+            while(p1 != m) {
+                ret[retP] = nums1[p1];
+                retP++;
+                p1++;
+            }
+        }
+        if (p2 != n) {  // 数组2有尾部
+            while(p2 != n) {
+                ret[retP] = nums2[p2];
+                retP++;
+                p2++;
+            }
+        }
+
+        System.arraycopy(ret, 0, nums1, 0, m + n);
+    }
+}