【leetcode】448. Find All Numbers Disappeared in an Array
题目如下:
解题思路:本题对时间复杂度和空间复杂度都有要求,特别是空间,所以不能用字典之类的来记录已经出现的值。这里可以采用值-下标映射的方法,即把所有元素移动到其值减1的对应的下标的位置上,移动完成后,下标和值不匹配的元素即为缺失的number。例如输入[4,3,2,7,8,2,3,1],
[4, 3, 2, 7, 8, 2, 3, 1] # 初始状态
[7, 3, 2, 4, 8, 2, 3, 1] #4和下标为4-1的元素互换
[3, 3, 2, 4, 8, 2, 7, 1] #7和下标为7-1的元素互换
[2, 3, 3, 4, 8, 2, 7, 1] #依次类推
[3, 2, 3, 4, 8, 2, 7, 1]
[3, 2, 3, 4, 8, 2, 7, 1]
[3, 2, 3, 4, 8, 2, 7, 1]
[3, 2, 3, 4, 8, 2, 7, 1]
[3, 2, 3, 4, 8, 2, 7, 1]
[3, 2, 3, 4, 1, 2, 7, 8]
[1, 2, 3, 4, 3, 2, 7, 8]
[1, 2, 3, 4, 3, 2, 7, 8]
[1, 2, 3, 4, 3, 2, 7, 8]
[1, 2, 3, 4, 3, 2, 7, 8]
[1, 2, 3, 4, 3, 2, 7, 8] #标红的元素和下标不匹配
代码如下:
class Solution(object): def findDisappearedNumbers(self, nums): """ :type nums: List[int] :rtype: List[int] """ inx = 0 while inx < len(nums): # [4,3,2,7,8,2,3,1] if nums[inx] != nums[nums[inx]-1]: #swap 4 and 7, to make val match inx src = nums[inx] # 4 desc = nums[nums[inx]-1] # 7 nums[inx] = desc nums[src - 1] = src else: inx += 1 res = [] for i,v in enumerate(nums): if i + 1 != v: res.append(i+1) return res