Leetcode 88. Merge Sorted Array (Easy)

Leetcode Top Interview 150 的其中一題

題目描述

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.

範例

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.

Constraints

• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -10^9 <= nums1[i], nums2[j] <= 10^9

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

解法

解法一

第一個解法有點簡單暴力，把 nums2 的值放到nums1 後段後，直接做排序

func merge(nums1 []int, m int, nums2 []int, n int)  {
    for j := m; j < len(nums1); j++ {
        nums1[j] = nums2[j-m]
    }

    sort.Ints(nums1)
}

解法二

這個解法很巧妙！因為 nums1 最後 n 格是空的，所以我們可以從 nums1 有數值的地方的尾巴(nums1最大的值) 和 nums2 最大的值開始比較，把比較大的填到 nums1 的最後，以此類推，從大的開始填

func merge(nums1 []int, m int, nums2 []int, n int)  {
    k := m+n-1 //最後一格
    i := m-1
    j := n-1

    for j>=0 {
        if i >= 0 && (nums1[i] > nums2[j]) {
            nums1[k] = nums1[i]
            i--
        } else {
            nums1[k] = nums2[j]
            j--
        }
        k--
    }
}