【题目】
合并两个数组,两个数组已分别顺序排好,不能使用额外数组,就存在nums1中
You are given two integer arrays nums1
and nums2
, sorted in non-decreasing order, and two integers m
and n
, representing the number of elements in nums1
and nums2
respectively.
Merge nums1
and nums2
into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1
. To accommodate this, nums1
has a length of m + n
, where the first m
elements denote the elements that should be merged, and the last n
elements are set to 0
and should be ignored. nums2
has a length of n
.
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] Explanation: The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
Example 2:
Input: nums1 = [1], m = 1, nums2 = [], n = 0 Output: [1] Explanation: The arrays we are merging are [1] and []. The result of the merge is [1].
Example 3:
Input: nums1 = [0], m = 0, nums2 = [1], n = 1 Output: [1] Explanation: The arrays we are merging are [] and [1]. The result of the merge is [1]. Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
【思路】
归并排序,讲的很清晰https://www.bilibili.com/video/BV1tT4y1w7hR
从后往前比较,
当P1P2都没走到头时,需要比较存储,
当P1走到头,P2还没时,直接存储P2
当P2走到头,P1还没时,直接返回(因为本来就往1里存,不会再变动)
【代码】
class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { int p1=m-1; int p2=n-1; int p=m+n-1; while(p1>=0&&p2>=0){ nums1[p--]=nums1[p1]>nums2[p2]?nums1[p1--]:nums2[p2--]; } while(p2>=0){ nums1[p--]=nums2[p2--]; } } }