合并排序数组Javascript

Question

我正在解决一些 LeetCode 的各种问题，作为我日常不会遇到的 JavaScript 概念的练习。

从简单的部分开始，我很困惑为什么这个合并数组不起作用？我发现我几乎从不拼接，因为我习惯于迭代并返回一个新元素，而且我很少使用需要直接修改的巨大数据集。

我的 Jasmine 错误如下

检查排序合并

    ✗ Array values are merged and sorted
      - Expected $.length = 6 to equal 3.
      Expected $[2] = 2 to equal 3.
      Unexpected $[3] = 3 in array.
      Unexpected $[4] = 5 in array.
      Unexpected $[5] = 6 in array.

下面是代码。

//////////////////
// INSTRUCTIONS //
//////////////////

// Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
// The number of elements initialized in nums1 and nums2 are m and n respectively.
// You may assume that nums1 has a size equal to m + n such that it has enough space to hold additional elements from nums2.

const nums1 = [1, 2, 3];
const m = 3;
const nums2 = [2, 5, 6];
const n = 3;

const mergeArray = (nums1, nums2) => {
  for (let index = 0; index < nums1.length - 1; index++) {
    if (nums2[index] >= nums1[index] && nums2[index] < nums1[index+1] ) {
      nums1.splice(index, 0, nums2[index]);
    }
  }
  return nums1;
};

module.exports = function () {
  describe("Check for Sorted Merge", () => {
    it("Array values are merged and sorted", () => {
      expect(nums1.concat(nums2).sort()).toEqual(mergeArray(nums1, nums2));
    });
  });
};

Answer 1

也许是这样的？由于两个数组都已排序，我们不需要将每个 m 元素与每个 n 元素进行比较。我们可以通过跟踪起点来节省一些时间复杂度，并在每次从 nums2 中找到 n 元素的位置时更新它。希望它有意义，但如果不是，我可以尝试更彻底地解释它。

const nums1 = [1, 2, 3]; // will always be sorted
const m = 3;
const nums2 = [2, 5, 6]; // will always be sorted
const n = 4;

const mergeArray = (nums1, nums2) => {
  let start_idx = 0;
  for (let num of nums2) {
    for (let idx = start_idx; idx < nums1.length; idx++){
      if (num <= nums1[idx]) {
        nums1.splice(idx, 0, num);
        start_idx = idx;
        break;
      }
      if (idx == nums1.length - 1){
        nums1.push(num);
        start_idx = idx;
        break;
      }
    }    
  }
  return nums1;
};

const res1 = nums1.concat(nums2).sort((a,b) => a-b);
const res2 = mergeArray(nums1, nums2);
console.log(res1, res2);

//JSON.stringify not ideal for arr/object comparison, but it works here for a quick check:
console.assert(JSON.stringify(res1) == JSON.stringify(res2), "sorts are not identical");