leetcode 2295. Replace Elements in an Array （python）

持续创作，加速成长！这是我参与「掘金日新计划 · 6 月更文挑战」的第17天，点击查看活动详情

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 operations[i][0] with operations[i][1].

It is guaranteed that in the ith operation:

operations[i][0] exists in nums. operations[i][1] 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].
复制代码

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].
复制代码

Note:

n == nums.length
m == operations.length
1 <= n, m <= 10^5
All the values of nums are distinct.
operations[i].length == 2
1 <= nums[i], operations[i][0], operations[i][1] <= 10^6
operations[i][0] will exist in nums when applying the ith operation.
operations[i][1] will not exist in nums when applying the ith operation.
复制代码

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 。对这个数组应用 m 个操作，在第 i 个操作中，将数字 operations[i][0] 替换为 operations[i][1] 。

题目保证在第 i 次操作中：

operation[i][0] 存在于 nums 中
operation[i][1] 在 nums 中不存在

返回应用所有操作后得到的数组。

这道题其实按照题意就可以了，需要注意的地方是我们要使用字典 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。

时间复杂度为 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[c] = i
	        for i in range(N):
	            x, y = operations[i]
	            idx = d[x]
	            nums[idx] = y
	            d[y] = idx
	        return nums	
复制代码

运行结果

80 / 80 test cases passed.
Status: Accepted
Runtime: 1826 ms
Memory Usage: 73.6 MB
复制代码

原题链接

leetcode.com/contest/wee…

您的支持是我最大的动力