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.
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 }
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 }
原文：http://www.cnblogs.com/ireneyanglan/p/4799827.html