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leetcode 2295. Replace Elements in an Array
描述
You are given a 0-indexed array nums that consists of n distinct positive integers. Apply m operations to this array, where in the ith operation you replace the number operationsi with operationsi.
It is guaranteed that in the ith operation:
operationsi exists in nums.
operationsi does not exist in nums.
Return the array obtained after applying all the operations.
Example 1:
Input: nums = [1,2,4,6], operations = [[1,3],[4,7],[6,1]]
Output: [3,2,7,1]
Explanation: We perform the following operations on nums:
- Replace the number 1 with 3. nums becomes [3,2,4,6].
- Replace the number 4 with 7. nums becomes [3,2,7,6].
- Replace the number 6 with 1. nums becomes [3,2,7,1].
We return the final array [3,2,7,1].
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Example 2:
Input: nums = [1,2], operations = [[1,3],[2,1],[3,2]]
Output: [2,1]
Explanation: We perform the following operations to nums: - Replace the number 1 with 3. nums becomes [3,2].
- Replace the number 2 with 1. nums becomes [3,1].
- Replace the number 3 with 2. nums becomes [2,1].
We return the array [2,1].
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Note:
n == nums.length
m == operations.length - <= n, m <= 10^5
All the values of nums are distinct.
operations[i].length == 2 - <= nums[i], operationsi, operationsi <= 10^6
operationsi will exist in nums when applying the ith operation.
operationsi will not exist in nums when applying the ith operation.
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解析
根据题意,给定一个由 n 个不同的正整数组成的索引为 0 的数组 nums。对这个数组利用 m 个操作,在第 i 个操作中,将数字 operationsi 替换为 operationsi。
题目保障在第 i 次操作中:
operationi 存在于 nums 中
operationi 在 nums 中不存在
返回利用所有操作后失去的数组。
这道题其实按照题意就可能了,需要注意的地方是咱们要使用字典 d 来保存每个元素的索引,并且在进行操作之后将每个字符对应的索引进行更新即可。
工夫复杂度为 O(N),空间复杂度为 O(N)。
解答
class Solution(object):
def arrayChange(self, nums, operations):
"""
:type nums: List[int]
:type operations: List[List[int]]
:rtype: List[int]
"""
N = len(operations)
d = {}
for i, c in enumerate(nums):
d = i
for i in range(N):
x, y = operations[i]
idx = d[x]
nums[idx] = y
d[y] = idx
return nums
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运行后果
80 / 80 test cases passed.
Status: Accepted
Runtime: 1826 ms
Memory Usage: 73.6 MB