88.Merge Sorted Array

Tags: [sort], [merge], [in_place]

Com: {fb}

Link: https://leetcode.com/problems/merge-sorted-array/#/description

Given two sorted integer arraysnums1andnums2, mergenums2intonums1as one sorted array.

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

Better Solution: In Place

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        index1 = m - 1
        index2 = n - 1
        main_index = m + n - 1

        while index1 >= 0 and index2 >= 0:
            n1 = nums1[index1]
            n2 = nums2[index2]
            if n1 == n2:
                nums1[main_index] = n1
                nums1[main_index - 1] = n2
                main_index -= 2
                index1 -= 1
                index2 -= 1
            elif n1 < n2:
                nums1[main_index] = n2
                main_index -= 1
                index2 -= 1
            else:
                nums1[main_index] = n1
                main_index -= 1
                index1 -= 1

        while index2 >= 0:
            nums1[main_index] = nums2[index2]
            main_index -= 1
            index2 -= 1

Revelation:

• Scan the nums1 and nums2 from right side to left side, to make use of the space of nums1.
• Don't forget to check the following part after the main while loop.

while index2 >= 0:
    nums1[main_index] = nums2[index2]
    main_index -= 1
    index2 -= 1

Note:

• Time complexity = O(n), n is the number of elements of the given array.
• Space complexity = O(1).

Solution:

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        result = []
        index_1 = 0
        index_2 = 0
        while index_1 < m and index_2 < n:
            n1 = nums1[index_1]
            n2 = nums2[index_2]
            if n1 == n2:
                result.append(n1)
                result.append(n2)
                index_1 += 1
                index_2 += 1
            elif n1 < n2:
                result.append(n1)
                index_1 += 1
            else:
                result.append(n2)
                index_2 += 1

        result += nums1[index_1:m]
        result += nums2[index_2:n]

        for i in xrange(len(result)):
            nums1[i] = result[i]

Note:

• Time complexity = O(m + n).
• Space complexity = O(m + n).