[Leetcode 88]合并有序数组Merge Sorted Array 归并排序

【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--];
        }
    }
}