技術標籤：LeetCode演算法與資料結構ArrayTwo PointersLeetcode

Topic

Array, Two Pointers

Description

https://leetcode.com/problems/merge-sorted-array/

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

Note:

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

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

Constraints:

• -10^9 <= nums1[i], nums2[i] <= 10^9
• nums1.length == m + n
• nums2.length == n

Analysis

參考歸併排序演算法中的歸併程式碼。

Submission

public class MergeSortedArray {

	//我寫的
	public void merge1(int[] nums1, int m, int[] nums2, int n) {
		for (int o = m-- + n-- - 1; m >= 0 || n >= 0; o--) {
			if (m < 0) {
				nums1[o] = nums2[n--];
			} else if (n < 0) {
				nums1[o] = nums1[m--];
			} else if (
 
nums1[m] > nums2[n]) {
				nums1[o] = nums1[m--];
			} else {// if(nums1[m] <= nums2[n]) {
				nums1[o] = nums2[n--];
			}
		}
	}

	//別人寫的
	public void merge2(int[] nums1, int m, int[] nums2, int n) {
		for (int o = m-- + n-- - 1; m >= 0 && n >= 0; o--)
			nums1[o] = (m < 0 || nums1[m] < nums2[n]) ? nums2[n--] : nums1[m--];
	}

}

Test

import static org.junit.Assert.*;

import java.util.Arrays;

import org.junit.Test;

public class MergeSortedArrayTest {

	@Test
	public void test() {
		MergeSortedArray obj = new MergeSortedArray();

		
		int[] nums1 = {1, 2, 3, 0, 0, 0};
		int[] nums2 = {2, 5, 6};
		
		int[] expected = {1, 2, 2, 3, 5, 6};
		
		obj.merge1(nums1, 3, nums2, 3);
		
		System.out.println(Arrays.toString(nums1));
		assertArrayEquals(expected, nums1);
		
	}
	
	@Test
	public void test2() {
		MergeSortedArray obj = new MergeSortedArray();
		
		
		int[] nums1 = {1, 2, 3, 0, 0, 0};
		int[] nums2 = {2, 5, 6};
		
		int[] expected = {1, 2, 2, 3, 5, 6};
		
		obj.merge2(nums1, 3, nums2, 3);
		
		System.out.println(Arrays.toString(nums1));
		assertArrayEquals(expected, nums1);
		
	}
}