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LeetCode 1122 数组的相对排序-简单-unordered_map容器的应用
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发布时间:2019-03-06

本文共 2250 字,大约阅读时间需要 7 分钟。

    
Relative Sort Arrays

Relative Sort Arrays Problem

Given two arrays, arr1 and arr2, where each element in arr2 is unique and appears in arr1, we need to sort arr1 such that the relative order of elements matches arr2. Elements not present in arr2 should be placed at the end in ascending order.

Example:

Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]

Output: [2,2,2,1,4,3,3,9,6,7,19]

Approach

To solve this problem, we can follow these steps:

  • Mapping Elements: First, create a mapping of each element in arr2 to its position. This helps in determining the required order of elements in arr1.
  • Sorting arr1: Sort the elements in arr1 based on their positions in arr2. Elements not present in arr2 are placed at the end in ascending order.

Implementation Steps

In code, we can implement this using a hash table (unordered_map) to store the positions of elements in arr2. Then, sort arr1 using a custom comparator that checks the positions from arr2. Finally, append any elements from arr1 that are not in arr2, sorted in ascending order.

Code Example

class Solution {public:    vector
relativeSortArray(vector
& arr1, vector
& arr2) { unordered_map
positionMap; // Populate the position map for (int i = 0; i < arr2.size(); ++i) { positionMap[arr2[i]] = i; } // Sort arr1 based on the position map sort(arr1.begin(), arr1.end(), [&positionMap](int a, int b) { return positionMap[a] < positionMap[b]; }); // Append elements from arr1 not present in arr2, sorted vector
extra; for (int num : arr1) { if (positionMap.find(num) == positionMap.end()) { extra.push_back(num); } } sort(extra.begin(), extra.end()); arr1.insert(arr1.end(), extra.begin(), extra.end()); return arr1; }

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