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10.01. Sorted Merge

Description

You are given two sorted arrays, A and B, where A has a large enough buffer at the end to hold B. Write a method to merge B into A in sorted order.

Initially the number of elements in A and B are m and n respectively.

Example:


Input:

A = [1,2,3,0,0,0], m = 3

B = [2,5,6],       n = 3



Output: [1,2,2,3,5,6]

Solutions

Solution 1: Two Pointers

We use two pointers $i$ and $j$ to point to the end of arrays $A$ and $B$ respectively, and a pointer $k$ to point to the end of array $A$. Then we traverse arrays $A$ and $B$ from back to front, each time putting the larger element into $A[k]$, then moving pointer $k$ and the pointer of the array with the larger element forward by one position.

The time complexity is $O(m + n)$, and the space complexity is $O(1)$.

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class Solution:
    def merge(self, A: List[int], m: int, B: List[int], n: int) -> None:
        i, j = m - 1, n - 1
        for k in reversed(range(m + n)):
            if j < 0 or i >= 0 and A[i] > B[j]:
                A[k] = A[i]
                i -= 1
            else:
                A[k] = B[j]
                j -= 1
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class Solution {
    public void merge(int[] A, int m, int[] B, int n) {
        int i = m - 1, j = n - 1;
        for (int k = A.length - 1; k >= 0; --k