Merge Sorted Array 解答

Question

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1and nums2 are m and n respectively.

Solution 1 Naive Way

Time complexity O(n), space cost O(n)

 1 public class Solution {
 2     public void merge(int[] nums1, int m, int[] nums2, int n) {
 3         int[] originNums1 = Arrays.copyOf(nums1, m);
 4         int pointer1 = 0, pointer2 = 0, i = 0;
 5         while (pointer1 < m && pointer2 < n) {
 6             if (originNums1[pointer1] <= nums2[pointer2]) {
 7                 nums1[i] = originNums1[pointer1];
 8                 pointer1++;
 9             } else {
10                 nums1[i] = nums2[pointer2];
11                 pointer2++;
12             }
13             i++;
14         }
15         while (pointer1 < m) {
16             nums1[i] = originNums1[pointer1];
17             pointer1++;
18             i++;
19         }
20         while (pointer2 < n) {
21             nums1[i] = nums2[pointer2];
22             pointer2++;
23             i++;
24         }
25     }
26 }

Solution 2

The key to solve this problem is moving element of A and B backwards. If B has some elements left after A is done, also need to handle that case.

Time complexity O(n), space cost O(1)

 1 public class Solution {
 2     public void merge(int[] nums1, int m, int[] nums2, int n) {
 3         while (m > 0 && n > 0) {
 4             if (nums1[m - 1] >= nums2[n - 1]) {
 5                 nums1[m + n - 1] = nums1[m - 1];
 6                 m--;
 7             } else {
 8                 nums1[m + n - 1] = nums2[n - 1];
 9                 n--;
10             }
11         }
12         while (n > 0) {
13             nums1[m + n - 1] = nums2[n - 1];
14             n--;
15         }
16     }
17 }
posted @ 2015-09-11 03:53  树獭君  阅读(183)  评论(0编辑  收藏  举报