LeetCode 88. 合并两个有序数组

88. Merge Sorted Array

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.

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?

解法思路：

1. 暴力法
2. 双指针
3. 逆向双指针

法一：

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        // voilent loop merge
        // Time: O((m+n)log(m+n))
        // Space: O(log(m+n))
        for (int i = 0; i < n; i++) {
            nums1[m + i] = nums2[i];
        }
        Arrays.sort(nums1);
    }
}

法二：

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        // double pointer
        // Time: O(m+n)
        // Space: O(m+n)
        int p1 = 0, p2 = 0, cur;
        int[] newArr = new int[m + n];
        while (p1 < m || p2 < n) {
            if (p1 == m) {
                cur = nums2[p2++];
            } else if (p2 == n) {
                cur = nums1[p1++];
            } else if (nums1[p1] < nums2[p2]) {
                cur = nums1[p1++];
            } else {
                cur = nums2[p2++];
            }
            newArr[p1 + p2 - 1] = cur;
        }
        for (int i = 0; i != m + n; ++i) {
            nums1[i] = newArr[i];
        }
    }
}

法三：

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        // reverse double pointer
        // Time: O(m+n)
        // Space: O(1)
        int p1 = m - 1, p2 = n - 1;
        int cur;
        int tail = m + n - 1;
        while (p1 >= 0 || p2 >= 0) {
            if (p1 == -1) {
                cur = nums2[p2--];
            } else if (p2 == -1) {
                cur = nums1[p1--];
            } else if (nums1[p1] > nums2[p2]) {
                cur = nums1[p1--];
            } else {
                cur = nums2[p2--];
            }
            nums1[tail--] = cur;
        }
    }
}

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