【LeetCode 0088】 【数组/双指针】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 [<u>1</u>,<u>2</u>,2,<u>3</u>,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?

Idea

根据约束条件，我们知道num1长度比num2长，逆向比较填充nums1数组
假设存m,n两个指针，分别指向nums1,numbs尾部
比较2个指针对应元素值的大小，不断将最大值移到nums1的后面
如果nums1[m] > nums2[n] ，将nums1[m]移动到 nums1的后面,m-1
如果nums1[m] < nums2[n] ，将nums2[n]移动到 nums1的后面,n-1
如果nums1的元素都比较完了，即m < 0,　只需要将 nums2剩下的元素依次搬进nums1
如果nums2的元素都比较完了，即n < 0,由于存在nums1中的元素已经有序，直接结束

JavaScript Solution

/**
 * @param {number[]} nums1
 * @param {number} m
 * @param {number[]} nums2
 * @param {number} n
 * @return {void} Do not return anything, modify nums1 in-place instead.
 */
var merge = function(nums1, m, nums2, n) {
    let last = m + n - 1;
    while(m > 0 || n > 0 ){
    //　只剩下nums2元素，依次搬入nums1的前面
        if( m < 1 ){
            nums1[last--] = nums2[n - 1]
            n --
        }
        // 只剩下nums1剩下的元素，直接结束流程
        if( n < 1 ){
            break;
        }
        // 比较大小，将大的元素搬到nums1后面
        if(nums1[m-1] > nums2[n-1]){
            nums1[last--] = nums1[m - 1]
            m -- ;
        }else{
            nums1[last--] = nums2[n - 1]
            n --;
        }
    }
    return nums1
};

原文地址：https://blog.csdn.net/avenccssddnn/article/details/140309302

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