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linyouxun avatar linyouxun commented on June 11, 2024
function OESort(nums) {
	var len = nums.length;
	nums.sort((i, i2) => i - i2);
	for(var i = 0; i < len - 1; i+=2) {
		var tmp = nums[i];
		nums[i] = nums[i + 1];
		nums[i + 1] = tmp; 
	}
	return nums
}

from fe-interview.

azl397985856 avatar azl397985856 commented on June 11, 2024
  1. 先排序,然后插入。 最容易想到,
    时间复杂度O(nlogn)
function OESort(list) {
    const res = [];
    let cnt = 0;

    list.sort((a, b) => a - b);

    for(let i = 1; i < list.length; i += 2) {
        cnt++;
        res[i] = list[(i / 2) >>> 0];
    }
    for(let i = 0; i < list.length; i += 2) {
        res[i] = list[cnt + (i / 2) >>> 0];
    }

    return res;
}

但是排序其实是没有必要的,于是有了下面的解法。

  1. 比较相邻,如果不满足条件则交换
    时间复杂度O(n)
function OESort(list) {
    function swap(l, i, j) {
      const t = l[i];
      l[i] = l[j];
      l[j] = t;
    }
    for (let i = 0; i < list.length; i += 2) {
      if (list[i] < list[i + 1]) {
        swap(list, i, i + 1);
      }
      if (i > 1 && list[i] < list[i - 1]) {
        swap(list, i, i - 1);
      }
    }

    return list;
  }

from fe-interview.

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